Concepts in Calculus II

Transcription

1 Concepts in Calculus II Beta Version UNIVERSITY PRESS OF FLORIDA Florida A&M University, Tallahassee Florida Atlantic University, Boca Raton Florida Gulf Coast University, Ft. Myers Florida International University, Miami Florida State University, Tallahassee New College of Florida, Sarasota University of Central Florida, Orlando University of Florida, Gainesville University of North Florida, Jacksonville University of South Florida, Tampa University of West Florida, Pensacola Orange Grove Texts Plus

2

3 Concepts in Calculus II Beta Version Miklós Bóna and Sergei Shabanov University of Florida Department of Mathematics University Press of Florida Gainesville Tallahassee Tampa Boca Raton Pensacola Orlando Miami Jacksonville Ft. Myers Sarasota

4 Copyright 202 by the University of Florida Board of Trustees on behalf of the University of Florida Department of Mathematics This work is licensed under a modified Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 Unported License. To view a copy of this license, visit creativecommons.org/licenses/by-nc-nd/3.0/. You are free to electronically copy, distribute, and transmit this work if you attribute authorship. However, all printing rights are reserved by the University Press of Florida ( Please contact UPF for information about how to obtain copies of the work for print distribution. You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). For any reuse or distribution, you must make clear to others the license terms of this work. Any of the above conditions can be waived if you get permission from the University Press of Florida. Nothing in this license impairs or restricts the author s moral rights. ISBN Orange Grove Texts Plus is an imprint of the University Press of Florida, which is the scholarly publishing agency for the State University System of Florida, comprising Florida A&M University, Florida Atlantic University, Florida Gulf Coast University, Florida International University, Florida State University, New College of Florida, University of Central Florida, University of Florida, University of North Florida, University of South Florida, and University of West Florida. University Press of Florida 5 Northwest 5th Street Gainesville, FL

5 Contents Chapter 6. Applications of Integration 36. The Area Between Curves 37. Volumes Cylindrical Shells Work and Hydrostatic Force Average Value of a Function 25 Chapter 7. Methods of Integration Integration by Parts Trigonometric Integrals Trigonometric Substitution Integrating Rational Functions Strategy of Integration Integration Using Tables and Software Packages Approximate Integration Improper Integrals 59 Chapter 8. Sequences and Series Infinite Sequences Special Sequences Series Series of Nonnegative Terms Comparison Tests Alternating Series Ratio and Root Tests Rearrangements Power Series Representation of Functions as Power Series Taylor Series 26 Chapter 9. Further Applications of Integration Arc Length Surface Area 39 Chapter and section numbering continues from the previous volume in the series, Concepts in Calculus I.

6 vi CONTENTS 62. Applications to Physics and Engineering Applications to Economics and the Life Sciences Probability 56 Chapter 0. Planar Curves Parametric Curves Calculus with Parametric Curves Polar Coordinates Parametric Curves: The Arc Length and Surface Area Areas and Arc Lengths in Polar Coordinates Conic Sections 96

7 CHAPTER 6 Applications of Integration 36. The Area Between Curves 36.. The Basic Problem. In the previous chapter, we learned that if the function f satisfies f(x) 0 for all real numbers x in the interval [a, b], then the area of the domain whose borders are the graph of f, the horizontal axis, and the vertical lines x = a and x = b is equal to b a f(x) dx. If there is no danger of confusion as to what a and b are, then this fact is sometimes informally expressed by the sentence the integral of f is equal to the area of the domain that is under the graph of f. What can we say about the area of the domain between two curves? There are several ways to ask this question. The easiest version, discussed by the following theorem, differs from the previous situation only in that the horizontal line is replaced by another function g. Theorem 6.. Let f and g be two functions such that, for all real numbers x [a, b], the inequality f(x) g(x) holds. Then the domain whose borders are the graph of f, the graph of g, and the vertical lines x = a and x = b has area A = b a (f(x) g(x)) dx. See Figure 6. for an illustration of the content of Theorem 6.. The reader is invited to explain why this theorem is a direct consequence of the fact that we recalled in the first paragraph of this section. The reader is also invited to explain why the theorem holds even if f and g take negative values. Example 6.. Compute the area A(D) of the domain D whose borders are the graph of the function f(x) =x 3 +, the function g(x) = x 2 +2, and the vertical lines x =2and x =3. See Figure 6.2 for an illustration of this specific example. Solution: In order to see that Theorem 6. is applicable, we must first show that, for all x [2, 3], the inequality f(x) g(x) holds.

8 2 6. APPLICATIONS OF INTEGRATION Figure 6.. Area enclosed by f(x) and g(x) between x = a and x = b. Figure 6.2. Area enclosed by the graphs of f(x) =x 3 + and g(x) =x between x = 2 and x =3. This is not difficult, since we only need to show that if x [2, 3], then f(x) g(x), that is, x 3 + x 2 +2, x 3 x 2, x 2 (x ), and this is clearly true since x 2, so x 2 4, and x, forcing x 2 (x ) 4.

9 36. THE AREA BETWEEN CURVES 3 Therefore, Theorem 6. applies, and we have 3 3 [ x A(D) = (x 3 +) (x 2 +2) dx = (x 3 x 2 4 ) dx = 4 x3 3 x 2 2 ] 3 = Intersecting Curves. Sometimes, the points a and b are determined by the curves themselves, and not given in advance. In that case, we have to compute them before we can apply Theorem 6.. Example 6.2. Find the area A(D) of the domain D whose borders are the graphs of the functions f(x) =x 2 +3x +5 and g(x) =2x 2 + 7x +8. See Figure 6.3 for an illustration. Solution: Let us find the points in which the graphs of f and g intersect. In these points, we have x 2 +3x +5=2x 2 +7x +8, 0=x 2 +4x +3, 0=(x + 3)(x +). That is, the two curves intersect in two points, and these points have horizontal coordinates a = 3 and b =. Furthermore, if x Figure 6.3. Area enclosed by the graphs of f(x) and g(x) between x = 3 and x =.

10 4 6. APPLICATIONS OF INTEGRATION [ 3, ], that is, if x is between those two intersection points, then f(x) g(x) =(x 2 +3x +5) (2x 2 +7x +8) = (x 2 +4x +3) = (x + 3)(x +) 0, since x +3 0 and x + 0. Therefore, if x [ 3, ], then f(x) g(x), and Theorem 6. applies. So we have A(D) = 3 (f(x) g(x)) dx = ] (x 2 +4x+3) dx = [ x3 3 2x2 3x 3 = 3. The situation becomes slightly more complicated if f g does not hold throughout the entire interval [a, b]. For instance, it could happen that f(x) g(x) at the beginning of the interval [a, b], and then, from a given point on, g(x) f(x). In that case, we split [a, b] up into smaller intervals so that on each of these smaller intervals, either f(x) g(x) or g(x) f(x) holds. Then we can apply Theorem 6. to each of these intervals. As on some of these intervals f(x) g(x) holds, while on some others g(x) f(x) holds, the application of Theorem 6. will sometimes involve the computation of (f(x) g(x)) dx and sometimes (g(x) f(x)) dx. The following theorem formalizes this idea. Theorem 6.2. Let f and g be two functions. Then the area of the domain whose borders are the graph of f, the graph of g, the vertical line x = a and the vertical line x = b is equal to b a f(x) g(x) dx. Note that Theorem 6. is a special case of Theorem 6.2, namely, the special case when f(x) g(x) = f(x) g(x) for all x [a, b]. Example 6.3. Let f(x) =x 3 +3x 2 +2x and let g(x) =x 3 + x 2. Compute the area A(D) of the domain whose borders are the graphs of f and g and the vertical lines x = 2 and x =. See Figure 6.4 for an illustration.

11 36. THE AREA BETWEEN CURVES 5 Figure 6.4. Graphs of f(x) and g(x) on[ 2, ]. Solution: In order to use Theorem 6.2, we need to compute f(x) g(x). We have f(x) g(x) =(x 3 +3x 2 +2x) (x 3 + x 2 ), 2x 2 +2x =0, 2x(x +)=0. That is, there are only two points where these two curves intersect, namely, at x = and x =0. Ifx orifx 0, then f(x) g(x) = 2x(x+) > 0, so f(x) g(x) = f(x) g(x) =2x 2 +2x. If <x<0, then f(x) g(x) < 0, so f(x) g(x) = g(x) f(x) = 2x 2 2x. Figure 6.5 shows the behavior of the function f(x) g(x). We can now directly apply Theorem 6.2. We get A(D) = = = [ 2x 3 = 3 + x2 f(x) g(x) dx (f(x) g(x)) dx + (f(x) g(x)) dx (2x 2 +2x) dx + (2x 2 +2x) dx ] (g(x) f(x)) dx ( 2x 2 2x) dx ] 0 [ ] [ 2x3 2x 3 3 x x2. 0

12 6 6. APPLICATIONS OF INTEGRATION Figure 6.5. Graph of f(x) g(x) on [ 2, ] Curves Failing the Vertical Line Test. Sometimes we want to compute the area between two curves that do not pass the vertical line test; that is, they contain two or more points on the same vertical line. Such curves are not graphs of functions of the variable x. If they pass the horizontal line test, that is, if they do not contain two points on the same horizontal line, then they can be viewed as functions of y. We can then change the roles of x and y in Theorems 6. and 6.2 and proceed as in the earlier examples of this section. Example 6.4. Compute the area A(D) of the domain between the vertical line x =4and the curve given by the equation y 2 = x. See Figure 6.6 for an illustration. Solution: Neither curve satisfies the vertical line test, but both satisfy the horizontal line test. Therefore, we set f(y) =4andg(y) =y 2.Itis Figure 6.6. Graph of y 2 = x and x =4.

13 37. VOLUMES 7 clear that the two curves intersect at the points given by y = 2 and y = 2. Between these two curves, the value of f(y) is larger. Therefore, Theorem 6. applies (with the roles of x and y reversed). So we have A(D) = = (f(x) g(x)) dx 4 y 2 dy = [4y y3 3 =6 6 3 ] 2 2 = Note that the geometric meaning of reversing the roles of x and y is simply reflecting all curves through the x = y line. That reflection does not change the area of any domain, so one can expect analogous methods of computing areas before and after that reflection Exercises. () Find the area of the domain whose borders are the vertical line x = 0, the vertical line x = 2, and the graphs of the functions f(x) =x and g(x) = sin x. (2) Find the area of the domain whose borders are the vertical line x =, the vertical line x = 3, and the graphs of the functions f(x) =x 3 and g(x) =e x. (3) Find the area of the domain between the graphs of the functions f(x) =x and g(x) =4x. (4) Find the area of the domain between the graphs of the functions f(x) =x 3 3x 2 and g(x) =x 2. (5) Find the area between the curves given by the equations x = 5y and x = y (6) Compute the area between the three curves f(x) =x, g(x) = x, and h(x) = Volumes 37.. Extending the Definition of Volumes. If a solid S can be built up using unit cubes, then we can simply say that the volume V (S) ofs is the number of unit cubes used to build S. However, if the borders of

14 8 6. APPLICATIONS OF INTEGRATION S are not planes, then this method will have to be modified. A ball or a cone is an example of this. So we would like to define the notion of volume so that it is applicable to a large class of solids, not just to those solids that are bordered by planes. This definition should agree with our intuition. It should also be in accordance with the fact that we can approximate all solids with a collection of very small cubes; therefore, V (S) must be close to the number of unit cubes used in the approximation. With these goals in mind, we recall that we already defined the area of a domain in the plane whose borders are the graphs of continuous functions. Building on that definition, we say that the volume of a prism is its base area times its height. More formally, let S be a solid whose base and cover are identical copies of the plate P, located at distance h from each other, on two parallel planes that are at distance h from each other. Then we define the volume of S to be V (S) =A(P )h, where A(P ) is the area of P. See Figure 6.7 for an illustration. In particular, the volume of a cylinder whose base is a circle of radius r and whose height is h is r 2 πh. Now let S be any solid located between the planes given by the equations x = a and x = b. In order to define and compute the volume V (S)ofS, we cut S into n parts by the planes x = x i for i =0,,...,n, where a = x 0 <x < <x n = b. Let n be large. Then the part S i of S that is between the planes x = x i and x = x i is well approximated by a prism P i described as follows. The height of P i is x =(b a)/n, the base plate and cover Figure 6.7. The volume of S is defined by V (S) =A(P )h.

15 37. VOLUMES 9 plate of P i are on the planes x = x i and x = x i, respectively, and the base and cover plates of P i are congruent to the intersection T i of S and the plane x = x i for some point x i [x i,x i ]. We can assume that x i is the midpoint of [x i,x i ], but that will turn out to be insignificant. Note that P i has volume A(T i ) x. If n is large, then the union of the prisms P i approximates S well, so V (S) should be defined in a way that assures that V (S) is close to (6.) n V (P i )= i= n A(T i ) x. i= As n goes to infinity, the Riemann sum on the right-hand side of (6.) has a limit. We define that limit to be the volume V (S) ofs, so V (S) = lim n n A(T i ) x. On the other hand, by the definition of the definite integral, we have lim n i= n A(T i ) x = i= b a A(t) dt, where A(t) is the area of the intersection of S and the plane x = t. Therefore, this integral is equal to the volume V (S) of the solid S. That proves the following theorem, which will be our main tool in this section. Theorem 6.3. Let S be a solid located between the planes x = a and x = b, and let A(t) be the area of the intersection of S and the plane x = t. Then the volume V (S) of S satisfies the equation V (S) = b a A(t) dt. Example 6.5. Compute the volume of the ball B whose center is at the origin and whose radius is r. Solution: For any given t [ r, r], the intersection of the plane x = t and the ball B is a circle. Let C t denote this circle. By the Pythagorean theorem, C t has radius r 2 t 2, and therefore, the area of C t is A(t) =(r 2 t 2 )π. See Figure 6.8 for an illustration.

16 0 6. APPLICATIONS OF INTEGRATION Figure 6.8. The volume of the ball B approximated by cylinders. Now we can use Theorem 6.3 to compute the volume of B. We get r V (B) = π ( r 2 π t 2) dt r [ = π r 2 t ] r 3 t3 r =2r 3 π 2 3 r3 π = 4 3 r3 π. There is nothing magical about the x axis as far as Theorem 6.3 is concerned. The argument that yielded that theorem can be repeated for the y axis instead of the x axis, yielding the following theorem. Theorem 6.4. Let S be a solid located between the planes y = a and y = b, and let B(t) be the area of the intersection of S and the plane y = t. Then the volume V (S) of S satisfies the equation V (S) = b a B(t) dt. Example 6.6. Let S be the right circular cone whose symmetry axis is the y axis, whose apex is at y = h, and whose base is a circle in the plane y =0with its center at the origin and with radius r. Find the volume of S. Solution: The cone S is between the planes y = 0 and y = h, and B(t) of Theorem 6.4 is easier to compute than A(t) of Theorem 6.3, so we use the former.

17 37. VOLUMES The intersection of the plane y = t and S is a circle. The radius r t of this circle, by similar triangles, satisfies r t r = h t h, showing that r t = r(h t)/h. Therefore, B(t) =r 2 (h t) 2 π/h 2, and Theorem 6.4 implies V (S) = h 0 = r2 π h 2 = r2 π h 2 = r2 π h 2 (r 2 (h t) 2 π/h 2 ) dt h 0 (h 2 2ht + t 2 ) dt [ h 2 t ht 2 + t3 3 h3 3 = 3 hr2 π. See Figure 6.9 for an illustration Annular Rings. In the examples that we have solved so far, the computation of A(t) or B(t), that is, the computation of the area of the intersection between a solid and a horizontal or vertical plane, was not difficult. That computation could be done directly. There are situations in which the domains whose areas we need to compute are not convex; that is, visually speaking, there is a hole in ] h 0 Figure 6.9. Right circular cone.

18 2 6. APPLICATIONS OF INTEGRATION them. This happens particularly often when S is obtained by rotating a domain D around a line. Example 6.7. Let D be the domain between the two curves y = x 2 = f(x) and y = 2x = g(x) and let S be the solid obtained by rotating D about the line x = 3. Find the volume of S. Solution: The two curves intersect at the points (0, 0) and (2, 4). The intersection of the horizontal plane y = t with S has the form of an annular ring, which is sometimes informally called a washer. This is simply a smaller circle cut out off the middle of a larger circle, so that the two circles are concentric. If the larger circle has radius r and the smaller circle has radius r 2, then the annular ring has area π(r 2 r 2 2). This general recipe enables us to compute B(t) in the example at hand. The points in D satisfy x [0, 2] and y [0, 4]. As 0 x 2, the inequality x 2 2x holds. So, for t [0, 4], the point P i =(t/2,t) on the graph of g(x) =y =2x is closer to the y axis than the point P o =( t, t) on the graph of y = x 2 = f(x). (It takes a larger value of x to get the same value t = y by f than it takes to get the same value by g.) So the outer circle of the annulus will be given by the rotated image of the curve of f (the parabola), and the inner circle of the annulus will be given by the rotated image of the curve of g (the straight line). In particular, for fixed t, the radii are obtained as the distance of P o (resp. P i ) from the axis of rotation, that is, the line x = 3. For the inner radius, this yields r 2 = 3 t 2 = t +6 2, while for the outer radius, this yields r = 3 t = t +3. Therefore, the area of the annular ring that is the intersection of S and the plane y = t is given by B(t) =π(r 2 r 2 2)=π ( ) ( ) 2 2 t +6 t +3 π = π (6t 0.5 t24 ) 2 2t.

19 37. VOLUMES 3 Now we can apply Theorem 6.4 to compute V (S). We obtain V (S) = B(t) dt = π 6t 0.5 t tdt ] = π [4t.5 t3 2 t2 dt Special Cases of Theorems 6.3 and 6.4. Note that the solids we discussed so far in this section could be obtained by rotating the graph of a function around an axis. Such volumes are called volumes of revolution. Indeed, the ball of Example 6.5 can be obtained by rotating the graph of the function f(x) = r 2 x 2 (a semicircle) about the x axis. The cone of Example 6.6 can be obtained by rotating the graph of the function f(x) = x h + r = y (a straight line) about the y axis. r For such solids, the areas A(t) and B(t) appearing in Theorems 6.3 and 6.4 are easy to compute, since the intersections appearing in those theorems will be circles or annular rings. If S is a solid obtained by rotating a curve about the x axis, then the intersection of the plane x = t and S is a circle of radius f(x), and hence A(t) =f(x) 2 π.ifsis a solid obtained by rotating the curve of the function g(y) =x about the y axis, then the intersection of S and the plane y = t is a circle of radius g(t), and so B(t) =g(t) 2 π. This yields the following special versions of Theorems 6.3 and 6.4. Theorem 6.5. Let S be a solid between the planes x = a and x = b obtained by rotating the graph of the function f(x) =y about the x axis. Then we have b V (S) = f(t) 2 π dt. a Theorem 6.6. Let S be a solid between the planes y = a and y = b obtained by rotating the graph of the function g(y) =x about the y axis. Then we have b V (S) = g(t) 2 π dt. a The exercises at the end of this section will provide further examples for the uses of these theorems.

20 4 6. APPLICATIONS OF INTEGRATION If the domain to be rotated does not include the entire area between the curve and and coordinate axis (for instance, because it is a domain between two curves), then we get annular rings, which we discussed in the last section A Solid Not Obtained by Revolution. While volumes of revolution are a very frequent application of Theorems 6.3 and 6.4, they are not the only applications of those theorems. Example 6.8. Let S be a pyramid whose base is a square of side length a and whose height is h. Compute the volume of S. Solution: The first step is to place S in a coordinate system so that Theorem 6.3 can be applied. Let us place the axis of S on the x axis of the coordinate system, so that the center of the base of S is at the origin and the cusp of S is at x = h. This does not completely determine the position of S, because S could still rotate around the x axis. However, such rotations will not change the value of A(t) for any t [0,h], and so they are insignificant for the computation of V (S). Now note that, for any t [0,h], the intersection of S and the plane x = t is a square of side length a(h t)/h (by similar triangles). See Figure 6.0. Pyramid.

21 37. VOLUMES 5 Figure 6.0 for an illustration. So A(t) =a 2 (h t) 2 /h 2, and Theorem 6.3 implies h a 2 (h t) 2 V (S) = dt 0 = a2 h 2 h 2 [ h 2 t ht 2 + t3 3 ] h 0 = a2 h The Big Picture. Note that the theorems presented in this section on integrals confirm our intuition that if a certain function measures a quantity, then, under the appropriate conditions, the integral of that function measures a quantity that is somehow in a space that is one dimension higher. For instance, we saw earlier that if f(x) measured the height of a curve at a given horizontal coordinate x, then, under the appropriate conditions, f(x) dx measured the area under the curve. So taking integrals meant moving from one dimension to two. In this section, the functions A(t) and B(t) measured areas of domains in a given plane, while A(t) dt and B(t) dt measured volumes. So taking integrals meant moving from two dimensions to three Exercises. () Compute the volume of the solid between the planes x = and x = obtained by rotating the graph of the function f(x) =x 2 = y about the y axis. (2) Compute the volume of the solid between the planes x = and x = obtained by rotating the graph of the function f(x) =x 2 = y about the x axis. (3) Compute the volume of the solid between the planes x =0 and x = π obtained by rotating the graph of the function f(x) = sin x = y about the x axis. (4) Compute the volume of the solid obtained by rotating the domain between the curves y = x and y = x about the line y = 2. (5) Compute the volume of the solid obtained by rotating the domain between the curves y = x 4 and y = x about the line x =.

22 6 6. APPLICATIONS OF INTEGRATION (6) Compute the volume of a regular tetrahedron of side length z. A regular tetrahedron is a solid that has four faces, each of which is a regular triangle. (7) Compute the volume of the solid between the planes x = 0 and x = obtained by rotating the curve f(x) =e x = y about the y axis. (8) Compute the volume of the solid between the planes x =0 and x = obtained by rotating the curve of the function f(x) =x( x) about the x axis. 38. Cylindrical Shells 38.. An Alternative Method to Compute Volumes. In principle, Theorems 6.3 and 6.4 are simple methods to compute the volumes of solids. In practice, however, the areas A(t) and B(t) that appear in these theorems may be difficult to explicitly evaluate. One situation in which these areas are often difficult to compute is when the solid in question is obtained by rotating a domain around some line; that is, it is a solid of revolution. As an example, let us try to compute the volume of the solid S obtained by rotating about the y axis the domain bordered by the lines x =0,x = 3, and y = 0 and the graph of the function f(x) =3x 3 x 4. If we try to solve this problem using Theorems 6.3 or 6.4, we run into difficulties, because A(t) and B(t) will not be easy to compute. For instance, if we wanted to use Theorem 6.4, then, in order to compute B(t), we would need to describe the intersection of S and the plane y = t. For this, we would have to find the x coordinates of the points of that intersection; that is, we would need to find all real numbers x [0, 3] for which y =3x 3 x 4 = t. This is a fourth-degree equation for x, which is very difficult and cumbersome to solve. If we wanted to use Theorem 6.3, then, in order to compute A(t), we would need to describe the intersection of the plane x = t and S, which is not straightforward to do. In situations like this, that is, when the application of Theorems 6.3 and 6.4 leads to technical difficulties, it often helps to use another method called the method of cylindrical shells. A cylindrical shell C is simply a cylinder C of which a smaller cylinder C 2 is removed, so that C and C 2 have the same symmetry axis. See Figure 6. for an illustration. If C 2 is just a little bit smaller than C, then C looks like a shell, explaining the name cylindrical shell.

23 38. CYLINDRICAL SHELLS 7 Figure 6.. Single cylindrical shell. If C and C 2 both have height h and C i has radius r i, then the volume of C can be computed as V (C) =V (C ) V (C 2 ) = hrπ 2 hr2π 2 = hπ(r 2 r2). 2 Note that the last form of V (C) can be rearranged as (6.2) V (C) =h (r r 2 ) 2π(r + r 2 ). 2 This way of writing V (C) might seem contrived at first sight. However, it has the following motivation. Note that r r 2 is the width of C, while h is its height. Finally, if we flatten C out in the plane, it will become a brick with side lengths h, r r 2, and, 2π r +r 2, since 2 the length of the missing side is equal to the circumference of a circle whose radius is the average of the radii of C and C 2. In other words, (6.2) says that the volume of a cylindrical shell is equal to the product of its height, width, and length (if the latter is interpreted properly). We are now in a position to use cylindrical shells to compute volumes. Let S be a solid that is obtained by rotating the domain D,

24 8 6. APPLICATIONS OF INTEGRATION which lies below the curve of f(x) =y from x = a to x = b, about the y axis. In order to estimate V (S), we cut [a, b] inton intervals of equal length using points a = x 0 <x < <x n <x n = b. For each integer i [,n], we will take a cylindrical shell S i, which will roughly cover the part of S that is obtained by rotating the part of D that is between the lines x = x i and x = x i about the y axis. More precisely, this shell will be obtained by removing the cylinder C i,2 from the cylinder C i,, where C i, and C i,2 are both cylinders whose symmetry axis is the y axis and whose height is f(x i ) for the midpoint x i of the interval [x i,x i ]. The radius of C i, is f(x i ), and the radius of C i,2 is f(x i ). If we set x =(b a)/n, then (6.2) implies that V (S i )=f(x i ) x 2πx i, since x i is the midpoint of the interval [x i,x i ]. Summing the last displayed equation over all i, weget (6.3) V (S) n f(x i ) x2πx i, i= since the union of the shells S i has roughly identical volume to S. As n goes to infinity, this approximation gets better and better, and the Riemann sum on the right-hand side of (6.3) converges to the corresponding definite integral. Hence, we have proved the following theorem. Theorem 6.7. The volume V (S) of the solid obtained by rotating the domain D whose borders are the curve of f(x) =y, the lines x = a and x = b, and the horizontal axis y =0about the y axis is equal to V (S) = b a 2πxf(x) dx. Example 6.9. Compute the volume of the solid S obtained by rotating about the y axis the domain bordered by the lines x =0, x =3, and y =0and the graph of the function f(x) =3x 3 x 4.

25 38. CYLINDRICAL SHELLS 9 Figure 6.2. (a) The curve of y =3x 3 x 4 and (b) the solid obtained by its rotation. Solution: By Theorem 6.7, we have V (S) =2π x(3x 3 x 4 ) dx =2π (3x 4 x 5 ) dx 0 [ ] 3x 5 3 =2π 5 x6 6 0 ( 729 =2π ) 6 = The axis around which we rotate a domain does not have to be a coordinate axis in order for the method of cylindrical shells to be applicable. We can apply the method as long as we can decompose the solid in question into cylindrical shells whose height and radius we can compute. Example 6.0. Let S be the solid obtained by rotating the domain whose borders are the horizontal axis, the vertical lines x = 0 and x =2, and the graph of the function f(x) =2x x 2 about the vertical line x =3. Compute V (S). Solution: We can decompose S into cylindrical shells whose center is on the vertical line x = 3. The shell containing the point x of the horizontal axis will have height f(x) =2x x 2 and radius 3 x.

26 20 6. APPLICATIONS OF INTEGRATION Figure 6.3. (a) The curve of y = x 2 /2 and (b) the solid obtained by its rotation. Therefore, we have V (S) = 2 2π(2x x 2 )(3 x) dx 0 2 ( =2π x 3 5x 2 +6x ) dx 0 [ ] x 4 2 =2π 4 5x3 3 +3x2 =2π Exercises. In () (3), use the method presented in this section to compute the volume of the solid obtained by rotating the domain between the given curves about the y axis. () f(x) =x 3 = y, x =,y =0. (2) f(x) = x 2, x =2,x =3,y =0. (3) f(x) =x, g(x) = x, x =2. In (4) (7), use the best available method to compute the volume of the solid obtained by rotating the domain between the given curves about the given axis. (4) f(x) =x = y, g(x) = x = y, x = 2, about x =.

27 39. WORK AND HYDROSTATIC FORCE 2 (5) f(x) =6 x, g(x) =x = y, y = 6, about the y axis. (6) f(x) =6 x, g(x) =x = y, y = 6, about the line y =6. (7) f(x) =6 x, g(x) =x = y, y = 6, about the line y = Work and Hydrostatic Force 39.. Work Moving a Point-like Object. In physics, the word work has a more specific meaning than in everyday life. Work in physics means that a force is extended to move an object a certain distance. The force F moving an object is computed by the formula F = m d2 s dt, 2 which is called Newton s second law. Here m is the mass of the object, while a = d2 s is its acceleration. So Newton s second law says that the dt 2 mass of an object is in direct proportion to the force needed to move it at constant acceleration. If a constant force F is exerted while an object moves distance d, then the work done by that constant force is computed by the formula W = Fd. Note that in the metric system, distance is measured in meters (m), time is measured in seconds (s), and therefore acceleration is measured in m/s 2. Mass is measured in kilograms (kg), so force is measured in kg m, which are called newtons (N), and, finally, work is measured in s 2 N m, which are also called joules (J). One joule is the work that is done when a force equal to newton moves an object a distance of meter. Example 6.. How much work is needed to lift a child of 20 kg to a height of 0.5 meters? Use the fact that gravitation causes downward acceleration of g =9.8 m/s 2. Solution: In order to lift the child, one needs to overcome the downward acceleration caused by gravity. This means that an upward force of m g =20kg 9.8 m s = 98 N 2 has to be exerted across a distance of d =0.5 meters. This yields W = Fd = 98 N 0.5 m=99j. So the work needed is 99 J. If the force exerted is not constant across the entire distance, but the distance can be split up into a few parts so that the force is constant on each part, then we can compute the work done by the force on each

28 22 6. APPLICATIONS OF INTEGRATION part just as in the previous example, and then we can add the obtained amounts to get the total amount of work done across the entire distance. If the force exerted changes according to a continuous function f(d), then we can approximate the work done using the idea of the previous paragraph and then use integration to compute the total work done by the force as follows. Let a and b be real numbers and let us assume that an object is moving from a to b, and the force moving the object at a given point x is equal to f(x), where f is a continuous function. In order to compute the work done across the entire distance, let us split the interval [a, b] into n equal intervals, using points a = x 0 <x < <x n <x n = b, and set x =(a b)/n. Then the work done by the force on the interval I i =[x i,x i ]isaboutf(x i ) x, where x i is some sample point in I i. Indeed, f is continuous, so if n is large and therefore I i is short, then f does not change much on that interval, so the shape of the domain under the curve of f and above I i is roughly a rectangle. This means that the total work done by the force on [a, b] is close to (6.4) n f(x i ) x. i= As n gets larger, this approximation gets better, and so we define the total work done by the force across the interval [a, b] asthelimit of the sum in (6.4) as n goes to infinity. On the other hand, that sum is a Riemann sum, so its limit, as n, is the definite integral b f(x) dx. a In other words, we have proved the following theorem. Theorem 6.8. Let a and b be real numbers. If an object is moved from a to b by a force that is equal to f(x) at point x, where f is continuous on [a, b], then the total work done by the force on [a, b] is W = b a f(x) dx. Example 6.2. The force needed to extend a given spring x centimeters over its natural length is given by the function f(x) =70x. How much work is needed to extend the spring 0 cm over its natural length?

29 39. WORK AND HYDROSTATIC FORCE 23 Solution: By Theorem 6.8, we have W = xdx = [35x 2 ] 0. 0 =2.25 J. So 2.25 J of work is needed to stretch the spring 0 cm over its natural length. We point out that the law of physics that says that the force needed to extend a spring by x units over its natural length is equal to kx is called Hooke s law, and k is called the spring constant Hydrostatic Force. Let us say that we want to pump water out of a tank that has the shape of the southern half of a ball of radius (m). How much work is needed to do that? This question is more complex than the previous one since deeper layers of the hemisphere are smaller, and water in those layers has to travel farther in order to reach the top of tank. Therefore, we will cut the tank up into small layers and estimate the amount of work needed to pump out each layer of water. Let x = 0 denote the bottom of the tank and let x = denote the center of the top circle of the tank. Cut the tank into i horizontal layers by planes that are at heights 0=x 0 <x < <x n =. Here x i x i =/n = xfor all i. Let L i denote the ith layer. The shape of this layer is close to a cylinder of height x. Each water particle in this layer has to be pumped at a distance roughly equal to z i, where z i isapointin[z i,z i ]. The square of the radius of the cylinder approximating L i is, by the Pythagorean theorem, ( z i ) 2 =2z i zi 2, and therefore the volume of L i is close to V i = x(2z i zi 2 )π. See Figure 6.4 for an illustration. One cubic meter of water has a mass of 000 kg, so the density of water is ρ = 000 kg/m 3. Therefore, the mass of the water in L i is close to m i = ρv i π. In order to pump this water out of the tank, the downward acceleration caused by gravitation, that is, m i g, has to be overcome, across a distance of z i. Therefore, the work needed to pump out the water in L i is about m i g( z i ), and the work needed to pump out all the water in the tank is approximated by n n (6.5) m i g( z i )=ρ gπ x(2z i zi 2 )( z i ). i= i=

30 24 6. APPLICATIONS OF INTEGRATION Figure 6.4. The tank in a coordinate system, and its layer at height z. As n grows, the expression displayed in (6.5) approximates the needed work better and better. We define the total work needed (to pump all the water out of the tank) to be the limit of the sum shown in (6.5) as n goes to infinity. As that sum is a Riemann sum, its limit, as n goes to infinity, is the definite integral W = ρgπ 0 (2x x 2 )( x) dx. As the integrand is a polynomial, this integral is very easy to compute. We get that W = ρgπ x 3 3x 2 +2xdx 0 [ ] x 4 = ρgπ 4 x3 + x 2 = ρgπ/4 = J. So it takes almost 7700 J of work to pump out all the water from the tank Exercises. () How much work is done when a book of mass 2 kg is lifted.5 meters from its original location? 0

31 40. AVERAGE VALUE OF A FUNCTION 25 (2) If it takes 0 J of work to lift an object 2 meters, how much work does it take to lift that object an additional 3 meters? (3) If it takes 20 J of work to stretch a spring 20 cm over its natural length, how much work does it take to stretch that spring an additional 5 cm? (4) How much work is needed to pump out all the water from a tank of the shape of an inverted cone of height 0 whose top circle has radius 2? (Length is measured in meters.) (5) How much work is needed to pump out all the water from a tank of the shape of a cylinder of height 20 whose base circle has radius 30? (6) How much work is needed to pump out all the water from a tank of the shape of an inverted pyramid of height 5 whose top plate is a square of side length 0? 40. Average Value of a Function The concept of average is a simple one as long as we take the average of a finite number of values, such as the average price of a house in a given neighborhood or the average daily high temperature in a given city in a given month. If a,a 2,...,a n are real numbers, then (6.6) A =(a + a a n )/n is their average. However, what can we say about the average value of a function over a given interval [a, b]? We will clearly need a new definition for that since there are infinitely many real numbers in [a, b], so summing all of them and then dividing their sum by the number of summands is not an option. Here is an intuitive way of extending the definition of average to the values taken by a function over an interval. It follows from (6.6) that A is the only real number with the property that if we replace each of a,a 2,...,a n by A, then the sum (a + a a n )doesnot change. This observation suggests the following definition. Definition 6.. Let f be a function such that b f(x) dx exists. a Then the average value of f on the interval [a, b] is the real number c = b a f(x) dx. b a Indeed, c is the only real number with the property that if we replace f by the constant function f(x) =c, then the integral b f(x) dx does a not change.

32 26 6. APPLICATIONS OF INTEGRATION A more systematic approach is the following. As we saw when we first learned about integrals, b f(x) dx can be approximated in the a following way. Split [a, b] inton equal intervals and choose a point x i in the ith such interval. Take a rectangle of height f(x i ) over the ith interval. The average value of the n values of f taken at the points x i is, of course, A n = f(x )+f(x 2 )+ + f(x n ). n On the other hand, the total area of the n rectangles we have just defined is R n = b a n (f(x )+f(x 2 )+ + f(x n )). Comparing the last two displayed equations, we see that (6.7) A n = R n b a. If n goes to infinity, then the n rectangles will approximate the domain under the graph of f, and so the right-hand side of (6.7) will converge b a f(x) dx to, while the left-hand side will approximate the average value b a of f on [a, b]. Example 6.3. What is the average value A of sin x on the interval [0,π]? Solution: We have A = π 0 sin xdx π = [ cos x]π 0 π = ( ) π = 2 π. See Figure 6.5 for an illustration. It is worth pointing out that a continuous function f will actually take its average value on each interval. This is the content of the following theorem. Theorem 6.9. Let f be a continuous function on [a, b] and let c be the average value of f on [a, b]. Then there exists a real number x [a, b] such that f(x) =c.

33 40. AVERAGE VALUE OF A FUNCTION 27 Figure 6.5. The average value of sin x on [0,π]. Proof. It suffices to show that if m is the minimum of f on [a, b] and M is the maximum of f on [a, b], then m c M, and our claim follows from the intermediate value theorem. We know that m(b a) b a f(x) dx M(b a) for obvious geometric reasons. Now divide all three terms by b a to get m c M. Example 6.4. There is a real number x [0,π] such that sin x = 2/π. Solution: This follows from the previous example and Theorem Exercises. () Find the average value of x n on [0, ]. (2) Find the average value of tan x on [0,π/4]. (3) Find the average value of ln x on [,e]. (4) Find the average value of e x on [0, ]. (5) What is larger, the average value of sin x or the average value of cos x, if both averages are taken on the interval [0,π/2]? (6) What is larger, the average value of sin x on [0,π] or the average value of sin x on [ 4π, 7π]? Can you find an answer that does not involve computation?

34

35 CHAPTER 7 Methods of Integration 4. Integration by Parts 4.. Method of Integration by Parts. Let u and v be two differentiable functions of the variable x. We used the simple product rule (7.) (uv) = u v + uv to compute the derivative of the product of these two functions. Is there a similar rule for computing the integral of the product of two functions? In general, the answer is no. There is no rule that provides the integral of the product of two functions that would work in every case. However, there are many cases in which a relatively simple way of reversing the product rule of differentiation will give us the answer we are trying to obtain. Indeed, integrating both sides of the product rule (7.) of differentiation (7.) with respect to x, we get the identity u(x)v(x) = (u (x)v(x)) dx + (u(x)v (x)) dx or, after rearrangement, (7.2) (u (x)v(x)) dx = u(x)v(x) (u(x)v (x)) dx. Formula (7.2) is very useful if we want to compute the integral of the product of two functions, one of which can play the role of u and the other one of which can play the role of v. If we can compute u, and (uv ), then formula (7.2) enables us to compute (u v) as well. If we cannot carry out one or both of these computations, then formula (7.2) will not help. Example 7.. Compute xe x dx. Solution: We set u (x) =e x and v(x) =x. Then formula (7.2) is easy to apply, since v(x) =x and v (x) =. Therefore, (7.2) implies that xe x dx = e x x e x dx = e x x e x + C = e x (x ) + C. 29

36 30 7. METHODS OF INTEGRATION The reader is encouraged to verify that the obtained solution is correct by computing the derivative of e x (x ) and checking that it is indeed equal to e x x. At this point, the reader may be asking how we knew that we needed to set u (x) =e x and v(x) =x, and not the other way around. The answer is that the other distribution of roles, that is, u (x) =x and v(x) =e x would not have helped. Indeed, if we had chosen u and v in that way, we would have needed to compute (uv ) dx = (e x x 2 )/2 dx. That would have been more complex than the original problem. We should always choose u and v so that (uv ) dx is easy to compute. That usually means selecting v so that it becomes simpler when differentiated, and to select u so that u does not get much more complex when integrated (or at least one of these two desirable outcomes occur). Example 7.2. Compute x cos xdx. Solution: We set u (x) = cos x and v(x) = x, which means that u(x) = sin x and v (x) =. So formula (7.2) implies x cos xdx= x sin x sin xdx= x sin x + cos x + C. The technique of integration we have just explained is called integration by parts Advanced Examples. Sometimes the integrand does not seem to be a product, but it can be transformed in to one. The following is a classic example. Example 7.3. Compute ln xdx. Solution: The crucial observation is that writing ln x = ln x helps. Let u (x) =andv(x) =lnx. Then u(x) =x and v (x) =/x, so, crucially, u(x)v (x) =. Therefore, formula (7.2) yields ln xdx= ln xdx= x ln x dx = x ln x x + C. Sometimes integration by parts leads to an equation or a system of equations that needs to be solved in order to get the solution to our problem. Example 7.4. Compute e x cos x.

37 4. INTEGRATION BY PARTS 3 Solution: We set u (x) =e x and v(x) = cos x. Then u(x) =e x and v (x) = sin x, and formula (7.2) yields (7.3) e x cos xdx= e x cos x + e x sin x dx. So we could solve our problem if we could compute the integral e x sin xdx. We can do that by applying the technique of integration by parts again, setting u (x) =e x and v(x) = sin x. We obtain (7.4) e x sin xdx= e x sin x e x cos x dx. Finally, note that (7.3) and (7.4) is a system of equations with unknowns e x cos xdx and e x cos xdx. We can solve this system, for instance, by adding these two equations and noting that e x sin x cancels. We get the equation e x cos xdx= e x (cos x + sin x) e x cos xdx or e x cos xdx= ex 2 (cos x + sin x)+c. Note that substituting the obtained expression for e x cos xdxinto (7.4), we get a formula for e x sin xdx, namely, e x sin xdx= ex (sin x cos x)+c Definite Integrals. If we evaluate both sides of formula (7.2) from a to b and we apply the fundamental theorem of calculus, we get the identity (7.5) b a (u v) dx =[uv] b a b a (uv ) dx. Example 7.5. Evaluate 2 ln xdx. Solution: As we saw in Example 7.3, we can set u(x) =x and v(x) = ln x. Then u (x) =,v (x) =/x, and formula (7.5) yields 2 ln x =[x ln x] 2 2 dx = 2 ln 2.

38 32 7. METHODS OF INTEGRATION 4.4. Exercises. () Compute x sin xdx. (2) Compute x 2 e x dx. (3) Compute x ln xdx. (4) Compute x 2 ln x. (5) Evaluate 2 x cos 2x. (6) Evaluate 0 x2 sin x. 42. Trigonometric Integrals 42.. Powers of sin and cos. In this section, we consider functions of the form f(x) = sin m x cos n x and discuss techniques for their integration. It seems natural to first consider the cases when m or n is 0, that is, when f is just a power of sin or cos. Even these special cases will break up into further subcases. The easiest subcase is when the exponents are even numbers. In that case, we can use the trigonometric identities (7.6) cos 2x = 2 cos 2 x and (7.7) sin 2x = 2 sin x cos x to eliminate high powers of trigonometric functions in the integrand. Example 7.6. Compute cos 4 xdx. Solution: Using (7.6), we get that cos 2 +cos 2x x =, and so 2 ( ) 2 + cos 2x cos 4 x = = cos 2x + + cos2 2x Applying (7.6) again, with 2x replacing x, we get that cos 2 +cos 4x 2x =, 2 so the previous displayed equation turns into cos 4 x = 3 cos 2x + cos 4x Having eliminated the powers of cos, the integration is easy to carry out as follows: ( ) 3 cos 4 cos 2x cos 4x xdx= + + dx = 3x 8 + sin 2x 4 + sin 4x 32 + C.

39 42. TRIGONOMETRIC INTEGRALS 33 The computation is more complex if the integrand is an odd power of sin or cos. In that case, we separate one factor and convert the rest into the other trigonometric function, using the rule cos 2 x+sin 2 x =. Example 7.7. Compute sin 3 xdx. Solution: We have sin 3 x = sin x sin 2 x = sin x ( cos 2 x) = sin x sin x cos 2 x. The advantage of this form is that it makes integration by substitution easy. Indeed, set u = cos x, then du/dx = sin x, and so sin x cos 2 xdx= u 2 du = u3 3 + C = cos3 x + C. 3 Comparing the two displayed equations of this solution and noting that sin xdx= cos x, weget sin 3 x = cos x + cos3 x + C. 3 The methods shown above can be used to compute the integral of products of powers of sin x and cos x. In other words, the method allows us to compute cos m x sin n xdx as shown below. Example 7.8. Compute cos 2 x sin 3 xdx. Solution: Just as in Example 7.7, we separate one sin x factor. This accomplishes two things. It allows us to convert the remaining even number of sin x factors to cos x factors, and it allows us to integrate by substitution. Indeed, cos 2 x sin 3 xdx= cos 2 x sin 2 x sin xdx = cos 2 x( cos 2 x) sin x = cos 2 x sin x cos 4 x sin x = u 2 du + u 4 du = u3 3 + u5 5 + C = cos3 x + cos5 x 3 5 where we used the substitution u = cos x. + C,

40 34 7. METHODS OF INTEGRATION We can always proceed this way if at least one of m and n in the integrand cos m x sin n x is odd. Indeed, in that case, after separating one factor from that odd power, an even power remains, and that can be converted to the other trigonometric function using the identity sin 2 x + cos 2 x =. If both m and n are even, then we can use that identity right away. Example 7.9. Compute cos 4 x sin 2 xdx. Solution: We have cos 4 x sin 2 xdx= cos 4 x( cos 2 x) dx = cos 4 cos 6 x dx. Now note that we computed cos 4 dx in Example 7.6. You are asked to compute cos 6 x in Exercise The difference of these two results then provides the solution of the present example Powers of tan and sec. When integrating a product of the form tan m x sec n x, we will use the identity sec 2 x = tan 2 x + and the differentiation rules (tan x) = sec 2 x and (sec x) = sec x tan x. There are two easy cases, namely, when m is odd (and n is at least ) and when n 2iseven. In the first case, that is, when m is odd and n, we separate one factor of tan x sec x and express the remaining factors in terms of sec x by the identity + sec 2 x = tan 2 x. Then we substitute u = sec x, which leads to du = tan x sec x. dx Example 7.0. Compute tan 3 x sec xdx. Solution: Following the strategy explained above, we have tan 3 x sec xdx= tan 2 x tan x sec xdx = ( + sec 2 x) tan x sec xdx = ( +u 2 ) du = u + u3 3 + C = sec x + sec3 x 3 + C.

41 42. TRIGONOMETRIC INTEGRALS 35 In the second case, that is, when n 2 is even, we separate one factor of sec 2 x, express the remaining factors in terms of tan x using the identity sec 2 x = + tan 2 x, and substitute u = tan x, which leads to du = dx sec2 x. Example 7.. Compute sec 4 x dx. Solution: We have sec 4 xdx= sec 2 x sec 2 xdx = ( + tan 2 x) sec 2 xdx =(+u 2 ) du = u + u3 3 + C = tan x + tan3 x + C. 3 If we are not in these two easy cases, then there is no recipe that will always work. We then need to have a separate strategy for each problem. We will show examples of that in Exercises and Some Other Trigonometric Integrals. If our goal is to compute integrals of the form cos mx sin nx dx, cos mx cos nx dx, and sin mx sin nx dx, then we can often make use of the following identities: (7.8) sin a cos b = 2 sin(a b)+ sin(a + b), 2 (7.9) cos a cos b = 2 cos(a b)+ cos(a + b), 2 (7.0) sin a sin b = 2 cos(a b) cos(a + b). 2 Example 7.2. Compute cos 3x cos 5xdx. Solution: Using (7.9) with a = 3x and b = 5x and noting that cos 2x = cos 2x, we get that cos 3x cos 5x = cos 2x + cos 8x, and 2 2 so ( cos 3x cos 5xdx= 2 cos 2x + ) cos 8x dx 2 = 4 sin 2x + sin 8x + C. 6

42 36 7. METHODS OF INTEGRATION Exercises. () Compute sin 3 xdx. (2) Compute cos 6 xdx. (3) Compute sin 3 x cos 2 xdx. (4) Compute tan 2 x sec 4 xdx. (5) Compute tan 3 x sec 5 xdx. (6) Compute tan 5 x by separating one factor of tan 2 x in the integrand and expressing it in terms of sec 2 x. (7) Compute sec 3 xdx using integration by parts, with u (x) = sec 2 x and v(x) = sec x. 43. Trigonometric Substitution 43.. Reversing the Technique of Substitutions. Let us assume that we want to compute the area of a circle by viewing one-fourth of that circle as the domain under a curve. Let r be the radius of the circle, and let us place the center of the circle at the origin. Then the northeastern quarter of the circle, shown in Figure 7., is just the domain under the graph of the function f(x) = r 2 x 2, where x ranges from 0 to r. In other words, we need to compute the integral r (7.) r2 x 2 dx. 0 In Chapter 5, we presented the technique of integration by substitution. This technique worked in situations when the best way to compute an integral was to define a simple function of x, such as y(x) =x 2, and then continue the integration in terms of that new variable y. In order to compute the integral in (7.), we use the reverse of the strategy mentioned in the previous paragraph. We define another variable y so that x is a simple function f of y. It is important to Figure 7.. The northeastern quadrant of the unit circle.

43 43. TRIGONOMETRIC SUBSTITUTION 37 define f and y so that f is one-to-one, since that assures that f(y) =x is equivalent to f (x) =y. In computing the integral in (7.), we can set x = r sin y. Then dx/dy = r cos y, and the limits of integration are y = 0 and y = π/2. This yields r 0 r2 x 2 dx = π/2 r 2 r 2 sin 2 yrcos ydy 0 π/2 = r 2 sin 2 y 0 π/2 = r 2 cos 2 ydy 0 = r 2 [ y + sin 2y 4 = r 2 π 4. ] π/2 Note that we could write cos y for sin 2 y, since 0 y π/2, and in that interval, cos y is nonnegative. We point out that by converting the indeterminate integral 2 y + sin 2y r back to a function of x, we get that 4 r2 ( x 2 dx = r2 x ) 4 sin + r2 r 2 x x 2. The result that we are going to compute in the next example will be useful in the next section, when we will learn a technique to integrate rational functions. Example 7.3. Compute the integral dx. (+x 2 ) 2 Solution: We use the substitution x = tan y. Then y = tan (x), and so dy/dx =/( + x 2 ), and hence dy = dx/( + x 2 ). This yields ( + x 2 ) dx = 2 = = = y 2 0 +x dy 2 + tan 2 y dy cos 2 ydy + sin 2y 4

44 38 7. METHODS OF INTEGRATION = y sin y cos y = 2 tan (x)+ 2 x x 2 +. The last step is justified since (7.2) x x 2 + = tan y + tan 2 y = tan y cos2 y = sin y cos y. Figure 7.2 illustrates this trigonometric argument. Example 7.4. Compute the integral x 2. Solution: The denominator reminds us of the trigonometric identity tan 2 y = sec 2 y, and so, if y [0,π/2), then tan y = sec 2 y. Therefore, we use the substitution x = sec y. Then dx/dy = tan y sec y. Hence, we have x2 = sec2 y dx = tan y dx tan y sec y = dy tan y = sec ydy =ln sec y + tan y =ln x + x 2. Figure 7.3 illustrates this trigonometric argument. Figure 7.2. Some expressions from (7.2).

45 43. TRIGONOMETRIC SUBSTITUTION 39 Figure 7.3. Some expressions occuring in the solution of Example Summary of the Most Frequently Used Trigonometric Substitutions. The three examples that we have seen so far in this section show the three most frequently used reverse substitutions. That is, (i) To compute r 2 x 2 dx, use the reverse substitution x = r sin y. (ii) To compute integrals involving (r 2 + x 2 ) under a root sign or in the denominator of a fraction, use the reverse substitution x = r tan y. (iii) To compute x 2 r 2 dx, use the reverse substitution x = r sec y. Finally, a word of caution. The availability of the method of reverse substitution does not mean that this method is always the best one to compute an integral that contains a square root sign. One of the following exercises can be solved by another method faster (and, no, we are not revealing which one) Exercises. () Use the method presented in this section to compute the area of an ellipse determined by the equation (2) Compute 4x 2 dx. (3) Compute +x 2 dx. (4) Compute x x 5 dx. (5) Compute x 2 dx. x 2 4 (6) Compute x 2 2xdx. x 2 a 2 + y2 b 2 =.

46 40 7. METHODS OF INTEGRATION 44. Integrating Rational Functions 44.. Introduction. Recall that a rational function is the ratio of two polynomials, such as R(x) = P (x) Q(x) = 3x +5 2x 2 +4x +9. Integrating rational functions is relatively simple, because most of these functions can be obtained as sums of even simpler functions. If the degree of P (x) is at least as large as the degree of Q(x), then we can divide P (x) byq(x), getting a polynomial as a quotient, and possibly a remainder. That is, if the degree of P is at least as large as the degree of Q, then there exist polynomials P (x) and P 2 (x) such that the degree of P 2 (x) isless than the degree of Q(x) and R(x) = P (x) Q(x) = P Q ( x)+p 2 (x) Q(x) = P (x)+ P 2(x) Q(x). As P (x) is a polynomial, it is easy to integrate. Therefore, the difficulty of integrating R(x) lies in integrating R 2(x), which is a rational Q(x) function whose denominator is of higher degree than its numerator. For this reason, in the rest of this section, we focus on integrating rational functions with that property, that is, when the degree of the denominator is higher than the degree of the numerator. Example 7.5. Let R(x) = x3 +2x+. Then dividing P (x) by Q(x) x 2 x+ using long division, we get P (x) =(x + )(x 2 x +)+2x, so R(x) = P (x) Q(x) = x ++ 2x x 2 x +, and integrating Q(x) boils down to integrating 2x. x 2 x Breaking Up the Denominator. In order to decide how to break up a rational function R(x) into the sum of simpler terms, we analyze the denominator Q(x) of R(x). A theorem in complex analysis, sometimes called the fundamental theorem of algebra, implies that if q(x) is a polynomial whose coefficients are real numbers, then q(x) can be written as a product of polynomials that are of degree or 2. This decomposition, or factorization, of Q(x) will determine the way in which we break up our rational function into the sum of simpler terms. There are several cases to distinguish, based on the factorization of Q(x).

47 44. INTEGRATING RATIONAL FUNCTIONS Distinct Linear Factors. The easiest case is when Q(x) factors into the product of polynomials of degree, and each of these terms occurs only once. Example 7.6. Compute dx. x 2 +3x+2 Solution: Note that x 2 +3x+2=(x+)(x+2). Using that observation, we are looking for real numbers A and B such that (7.3) x 2 +3x +2 = A x + + B x +2 as functions, that is, such that (7.3) holds for all real numbers x. Multiplying both sides by x 2 +3x +2,weget (7.4) = A(x +2)+B(x +). If (7.4) holds for all real numbers x, it must hold for x = and x = 2 as well. However, if x =, then (7.4) reduces to = A, and if x = 2, then (7.4) reduces to = B. So we conclude that A = and B = are the numbers we wanted to find. It is now easy to compute the requested integral as follows: x 2 +3x +2 dx = x + dx x +2 dx = ln(x +) ln(x +2)+C. The above method can always be applied if Q(x) factors into a product of linear polynomials, each of which occurs only once. In particular, if Q(x) decomposes as a(x a )(x a 2 ) (x a k ), then we can decompose R(x) into a sum of the form A + A A k. x a x a 2 x a k After determining the numbers A i, we can integrate each of the above k summands Repeated Linear Factors. The next case is when Q(x) factors into linear terms, but some of these terms occur more than once. Example 7.7. Compute 2x+7 dx. (x+) 2 (x ) Solution: Just as in the previous case, we decompose the integrand into a sum of simpler fractions. We are looking for real numbers A, B, and C such that 2x +7 (x +) 2 (x ) = A x + + B (x +) + C 2 x.

48 42 7. METHODS OF INTEGRATION Multiplying both sides by the denominator of the left-hand side, we get 2x +7=A(x + )(x ) + B(x ) + C(x +) 2. Substituting x = in the last displayed equation yields 9 = 4C, so C =2.25. Substituting x = yields 5 = 2B, sob = 2.5. Finally, the coefficient of x 2 on the left-hand side is 0, while on the right-hand side, it is A + C. SoA + C = 0, yielding A = Now we are in a position to compute the requested integral. 2x +7 (x +) 2 (x ) = dx + dx + x + (x +) 2 x dx 2.25 = x + dx (x +) dx x dx = 2.25 ln(x +) ln(x ). x + In general, if a term (x+a) k occurs in Q(x), then the partial fraction decomposition of R(x) will contain one term with denominator (x+a) i for each i {, 2,...,k}. For instance, if Q(x) =(x+2) 3 (x+5) 2 (x 0), then R(x) will have a partial fraction decomposition of the form A x +2 + A 2 (x +2) + A 3 2 (x +2) + A 4 3 x +5 + A 5 (x +5) + A 6 2 x Distinct Quadratic Factors. The third case is when the factorization of Q(x) contains some quadratic factors that are irreducible (i.e., they are not the product of two linear polynomials with real coefficients), but none of these irreducible quadratic factors occurs more than once. In that case, after obtaining the partial fraction decomposition of R(x), we may have to resort to the formulas x 2 + = tan x + C and x 2 + a = ( x ) 2 a tan + C. a Example 7.8. Compute the integral 4x+2 dx. x 3 +x 2 +x+ Solution: It is easy to notice that setting x = turns the denominator to 0; hence, the denominator is divisible by x +. Dividing the denominator by x +, we get x 2 +, so the denominator factors as (x + )(x 2 + ). The factor x 2 + is irreducible (it is not divisible by

49 44. INTEGRATING RATIONAL FUNCTIONS 43 x b for any real number b, since no real number b satisfies the equation b 2 + = 0). Therefore, we are looking for real numbers A, B, and C such that 4x +2 (7.5) x 3 + x 2 + x + = A x + + B x Cx x 2 +. The reader is invited to verify that the third summand of the righthand side is necessary; that is, if the summand Cx is removed, then x 2 + no pair of real numbers (A, B) will satisfy (7.5). In order to find the correct values of A, B, and C, multiply both sides of (7.5) by (x +) (x 2 + ) and rearrange, to get 4x +2=(A + C)x 2 +(B + C)x + A + B. The coefficient of x 2 is 0 on the left-hand side, so it has to be 0 on the right-hand side. Therefore, A + C = 0. Similarly, the coefficient of x is 4 on the left-hand side, so it has to be 4 on the right-hand side, forcing B + C = 4. Similarly, the constant terms of the two sides have to be equal, and, consequently, A + B = 2. Solving this system of equations, we get A =, B = 3, and C =. Therefore, 4x +2 x 3 + x 2 + x + dx = dx +3 x + x 2 + dx + = ln(x +)+3tan x + 2 ln(x2 +). x x 2 + dx In general, if x 2 +ax+b is a quadratic factor in Q(x), then the partial fraction decomposition will contain a summand of the form E and x 2 +ax+b a summand of the form Fx. Again, the latter is necessary, since x 2 +ax+b a rational fraction of the form E+Fx will not equal one of the form x 2 +ax+b for any choice of E if F 0. E x 2 +ax+b Repeated Quadratic Factors. Finally, it can happen that the factorization Q(x) contains irreducible quadratic factors, some of which occur more than once. Example 7.9. Compute the integral x 3 +2x 2 +3x+7dx. x 4 +2x 2 + Solution: It is easy to see that the denominator factors as (x 2 +) 2. Hence, we are looking for real numbers A, B, C, and D such that x 3 +2x 2 +3x +7 x 4 +2x 2 + = A x Bx x C (x 2 +) 2 + Dx (x 2 +) 2. Multiplying both sides by x 4 +2x 2 + and rearranging, we get x 3 +2x 2 +3x +7=Bx 3 + Ax 2 +(B + D)x +(A + C).

50 44 7. METHODS OF INTEGRATION For each k, the coefficients of x k must be the same on both sides. Hence, A = 2 and B =,soc = 5 and D =2. Now we can compute the requested integral using the preceding partial fraction decomposition as follows: x 3 +2x 2 +3x +7 dx x 4 +2x 2 + ( 2 = x x x (x 2 +) + 2x 2 (x 2 +) 2 ) dx =2 tan x + 5x 2 ln(x2 +)+ 2(x 2 +) tan x x 2 + = 2 ln(x2 +)+4.5 tan x + 2 5x 2 x 2 +. Here we used the formula for that we computed in the last (x 2 +) 2 section, in Example 7.3. By now, the reader must know what the general version of the technique of the preceding example is. If the factorization of Q(x) contains (x 2 + ax + b) k, then, for each integer i such that i k, the partial fraction decomposition of R(x) will contain a summand of E the form i F and a summand of the form i x. (x 2 +ax+b) i (x 2 +ax+b) i Rationalizing Substitutions. There are situations when a function that is not a rational function can be turned into one by an appropriate substitution, and then it can be integrated by the methods presented in this section. The most frequent scenario in which this happens is when the integrand contains roots, but if those roots are replaced by another variable, we get a rational function in that other variable. Example Compute x x+ dx. Solution: We use the substitution x = y. Then dy/dx = 2 =. x 2y This leads to x dx = x + = = y y + 2ydy 2y 2 y + dy 2(y ) + 2 y + dy

51 45. STRATEGY OF INTEGRATION 45 = y 2 2y + 2 ln(y +) = x 2 x + 2 ln( x +). Note that the computation would have been very similar if the integrand contained some other root of x. Indeed, if the integrand contained r x instead of x, then we would have substituted y = r x, and that would have turned the integrand into a rational function of y. Indeed, y = x /r implies dy dx = x(/r), r dy dx = y r r, and therefore dx = ry r dy. In other words, dx is a equal to dy times a polynomial function of y, so, indeed, the integrand will be a rational function of y Exercises. () Compute 5 dx. x 2 +5x+4 (2) Compute 3 (3) Compute dx. x 4 +5x 2 +4 (4) Compute dx. x 4 +4x 2 +4 (5) Compute x 2 +4 dx. x 3 x (6) Compute x+3 dx. x 3 + (7) Compute x+ x 3 dx. (8) Compute 3 x 3 x+2 dx. x 3 +x 2 x dx. 45. Strategy of Integration We presented various integration techniques in this chapter and in some preceding chapters. The most general ones were integration by parts and integration by substitution. The most frequently studied special cases were related to trigonometric functions and their inverses. Reverse substitution came up in some special cases. We also discussed the integration of rational functions, using the technique of partial fractions. In short, we have learned a decent number of methods. For this very reason, it is sometimes not obvious which method we should use

52 46 7. METHODS OF INTEGRATION when trying to integrate a function. While there is no general rule, in this section we will provide a few guidelines. So let f be a function that is not equal to one of the functions whose integral we either know offhand or have a deterministic method to compute. That is, f is not a polynomial, f is not a rational function, f is not the function f(x) =a x or f(x) = log a x for some positive real number a, and f is not one of the basic trigonometric functions like sin x or tan x. Let us also assume that simple algebra will not help, that is, that f cannot be transformed into one of these elementary functions by simple algebraic transformations. Then how do we decide which method to use? 45.. Substitution. The method that needs the least amount of work, when it is available, is a simple substitution, so it is reasonable to try to use that method first. There is a particularly good chance for this approach to work when f is the composition of two functions, one of which has a constant derivative, or when f is of the form f(x) = h (g(x))g (x), since then f(x) = dh dx (x), and so f(x) dx = h(x). In the language of substitutions, this means that substituting y = g(x) will work, since fdx= h (g(x))g (x) dx = h (y) dy dx dx = h (y) dy. In other words, the integral of the composite function f is turned into something simpler, the integral of the function h. Example 7.2. Let f(x) = sin 2x. Then, using the substitution y =2x, we get that sin 2xdx= 2 sin ydy= cos y = cos 2x. 2 2 Example Let f(x) = 3x. Then we set y = x 2 + x2 +,so dy/dx =2x. This leads to 3x 3x x 2 + dx = y dx = 3 2 y dy = 3 2 ln y = 3 2 ln(x2 +). The reader should compute the integral sin n x cos xdx at this point.

53 45. STRATEGY OF INTEGRATION Integration by Parts. If f is the product of two functions, but substitution does not seem to help, then integration by parts is the logical next step. This technique is particularly useful when one of the two functions whose product is f is made significantly simpler by differentiation. Example Compute the integral xe x dx. Solution: Considering the integrand, we notice that substitution is unlikely to help, since x and e x are not closely related. On the other hand, the integrand is a product, and one of the terms, x, is made simpler by differentiation. Therefore, we choose the technique of integration by parts, with x = u and v = e x. Then u =, while v = e x, and we get xe x dx = xe x e x dx = xe x + e x =( x)e x Radicals. If the two most general methods (substitution and integration by parts) are not helpful, then it is quite possible that there is a root sign in the integrand. In that case, there are two specific methods that we can try, reverse substitution and rationalizing substitution. The easiest way to know when to use each of these two methods is to remember the relatively few cases in which reverse substitution works directly. As we have seen, these are the integrals involving, r 2 +x 2 r2 x 2, and x 2 r 2. If the integrand does not contain any of these functions, then it may be simpler to use a rationalizing substitution, such as in computing the integral x x+4 dx. The exercises at the end of this section will ask the reader to decide which method to use for a few specific examples If Everything Else Fails. If none of our methods work, then it may be that an unexpected transformation of the integrand may help, at least with relating the integral to one that is not quite as challenging to compute. Example Compute sin 4 x tan 2 x cos2 2xdx.

54 48 7. METHODS OF INTEGRATION Solution: We use the trigonometric identity tan x = sin x/ cos x, to rewrite the integrand. We get sin 4 x tan 2 x cos2 2xdx= sin 2 x cos 2 x cos 2 2xdx = 4 sin2 2x cos 2 2xdx = 6 sin2 4x dx, which is easy to integrate with the method we learned for powers of trigonometric functions. It is important to point out that sometimes there is indeed no solution; that is, there exist elementary functions f such that there is no elementary function F (x) satisfying F (x) =f(x). Examples of such functions f include e x2, e x /x, and. ln x Exercises. In all of the exercises below, compute the integral. () x 2 e 5x dx. (2) x dx. x 2 (3) x+ dx. x 3 + (4) e sin x cos xdx. (5) e x dx. e x (6) dx. x 2 +4 (7) x+4 dx. (8) dx. x ln x (9) dx. x ln x ln(ln x) 46. Integration Using Tables and Software Packages 46.. Tables of Integrals. Tables of integrals can be found in many calculus textbooks and on the Internet. The website is a good example, and we will use it as a reference in this subsection. (When we say the table of integrals, we mean that table.) No matter how extensive a table of integrals is, it cannot contain all integrals. It is therefore important to know how to use these tables to compute integrals that are not contained in the tables in the same form. The easiest case is when the integral to be computed is a special case of a more general integral that is in the table.

55 46. INTEGRATION USING TABLES AND SOFTWARE PACKAGES 49 Example Use the table of integrals to compute ln(x 2 +9) dx. Solution: Looking at the table of integrals found at the website we find that the integrand is a special case of integral (45), with a = 3. Using the formula given in the table for the general case with a = 3, we get the result ( x ) ln(x 2 +9)dx = x ln(x 2 +9)+6tan 2x + C. 3 Sometimes, we have to resort to integration by substitution to be able to use the table of integrals. Example Use the table of integrals to compute 9 4x 2 dx. Solution: Taking a look at the table of integrals, we find that formula (30) provides a formula for a 2 x 2 dx. In order to be able to use that formula, we set y =2x, which implies dy/dx = 2. Therefore, 9 4x2 dx = 9 y2 dy. 2 Now we can apply formula (30) from the table of integrals, to get ( ) 9 y2 dy = y 9 y2 + 9 y tan + C 9 y 2 = x 9 4x2 + 9 ( ) 2x 2 2 tan + C. 9 4x 2 Sometimes we need to carry out some algebraic manipulation before we can use the technique of substitutions in connection with using the table of integrals. Example Use the table of integrals to compute x 2 +2x+ dx. x 2 +2x+0 Solution: There is no integral in the table of integrals that would immediately stand out as one that is very similar to this one. The crucial observation is that the substitution y = x+ significantly simplifies our integrand. That substitution leads to the integral y y 2 dy, which, in turn, can be directly found in the table of integrals as item (36). Substituting x back into the obtained formula, we get x 2 +2x + x2 +2x +0 dx x2 +2x +0 = 2 9 x 2 ln ++ x2 +2x +0 + C.

56 50 7. METHODS OF INTEGRATION Software Packages. Computer software packages such as Maple and Mathematica are very useful tools of integration. These packages will compute the definite or indefinite integrals of a large class of functions, then they will present the results in a form that is usually, but not always, in the form the user expected. In this section, we show a few examples of these unexpected results and explain how to interpret them. To start with a very basic example, type int(xˆ2+3x,x); into Maple. We get the answer 3 x x2. This is the correct answer having constant term 0. Experimenting with other functions, we note that Maple always answers in this way, that is, without the constant C at the end. It is important not to forget this if we will be using the obtained function in some further computation. Maple does not always provide the simplest form for the integral that it computes. For instance, if we ask Maple to compute the indefinite integral x(x 2 +3) 6 dx, we get the output (7.6) 4 x x x x x x x2. However, it is very easy to compute x(x 2 +3) 6 dx by hand, using the substitution u = x 2. That substitution leads to the same solution, but in a much simpler form, namely, 4 (x2 +3) 7. Ifwewant to verify that this result indeed agrees with the one given by Maple and displayed in (7.6), we can ask Maple to expand the expression H = 4 (x2 +3) 7 using the expand command. We see that the expanded expression indeed agrees with the one given in (7.6), up to the constant terms at the end. There are other commands like expand that, are useful if we want to transform the output of an integration software package. The commands rationalize and simplify are examples of these. There is often more than one way to express integrals involving hyperbolic functions. A striking example is the following. If we ask Maple to compute dx by typing x 2 int(/(-xˆ2),x); Maple returns the answer tanh x. This is very surprising, since it is not difficult to integrate the integrand as a rational function, with no

57 46. INTEGRATION USING TABLES AND SOFTWARE PACKAGES 5 hyperbolic functions involved. Indeed, x = x x, and so (7.7) x dx = 2 2 ln(x +) x + ln(x )=ln 2 x. The result obtained by Maple actually agrees with the result given in (7.7), even if that is not obvious. Indeed, if y = tanh x, then, by definition, x = (e y e y )/(e y + e y ). Solving this equation for x+ y is not completely trivial, but at the end, it yields y = ln. x (Hint: Multiply both the numerator and the denominator by e y to get x =(e 2y )/(e 2y + ).) Finally, when computing definite integrals, Maple sometimes answers by using the acronyms of some rare functions. For instance, if we want to compute sin 0 (x2 ) dx by typing int(exp(sin(xˆ2)),x=0..). then we get the answer (/2)*FresnelS(sqrt(2)/sqrt(Pi))*sqrt(2)*sqrt(Pi). Here FresnelS refers to the Fresnel sine integral, a concept beyond the scope of this book. If we simply want to know a numerical value for 0 ex2 dx, wecantype evalf(int(sin(xˆ2),x=0..)) instead. Maple outputs as the answer Exercises. () Use the table of integrals to compute x ln(3x +5)dx. (2) Use the table of integrals to compute x cos ( π x) dx. 2 (3) Use the table of integrals to compute x+7 dx. x + 4x+5 (4) Use the Table of Integrals to compute x 2 dx. x 6 +6 (5) Use your favorite software package to compute xe x dx, x 2 e x dx, and x 3 e x dx. Do you see a pattern? Try to guess what x 4 e x is, then verify your guess by using your software package again. (6) Let f(x) = x 2 and let g(x) =. Clearly, f(x) = x 2 g(x) for all x where these functions are defined. Compute the integral of both functions with Maple. If the results seem different, show that they are in fact equal.

58 52 7. METHODS OF INTEGRATION 47. Approximate Integration Sometimes it is not possible to find the exact value of a definite integral b f(x) dx. It could happen that we cannot find the antiderivative of f(x) or that the antiderivative of f(x) is not an elementary a function. Or it could happen that f itself is not given by a formula, but instead, f is given by its graph, which is plotted by a computer program. In this case, we resort to methods of approximate integration Basic Approximation Methods. The key observation behind the approximation methods is the fact that if f(x) 0 for x [a, b], then b f(x) dx is equal to the area below the graph of the function f on a that interval. More precisely, b f(x) dx is equal to the area of the a domain bordered by the horizontal axis, the vertical lines x = a and x = b, and the graph of f. In order to estimate the area of this domain D, we cut D into small vertical strips. To do so, we choose real numbers a = x 0 <x < <x n = b and then consider the vertical lines x = x i for all i. There are several ways in which we can estimate the area of each strip. Let S i be the strip bordered by the lines x = x i and x = x i+. We could say that the area of S i is roughly equal to the area of the rectangle whose base is the interval [x i,x i ] and whose height is f(x i ). This is called the left-endpoint approximation. Or we could say that the area of S i is roughly equal to the area of the rectangle whose base is the interval [x i,x i ] and whose height is f(x i ). This is called the right-endpoint approximation. Or we could use f( x i ), the value that f takes at the midpoint x i of the interval [x i,x i ]. This is called the midpoint approximation. Summing over the allowed values of i (i.e., i ranges from to n), we see that these three methods provide the following three estimates for b f(x) dx. a (i) The left-endpoint approximation shows b n (7.8) f(x) dx (x i x i )f(x i ). a i= (ii) The right-endpoint approximation shows b n f(x) dx (x i x i )f(x i ). a i=

59 47. APPROXIMATE INTEGRATION 53 (iii) The midpoint approximation shows b n f(x) dx (x i x i )f( x i ). a i= As the above formulas suggest, it will be particularly easy to work with these formulas if the points x,x 2,...,x n split the interval [a, b] into equal parts, since in that case x i x i =(b a)/n for all i. Example Find the approximate value of 0 ex2 dx. Solution: Let us use the left-endpoint method with n = 4 and x, x 2, and x 3 splitting [0, ] into four equal parts. That means that x =/4, x 2 =/2, and x 3 =3/4. Then (7.8) implies e x2 dx ( e 0 + e /6 + e /4 + e 9/6).2759, 0 4 since x i x i =/4 for all i. If we use the right-endpoint method, with the same set of points x i,weget e x2 dx ( e /6 + e /4 + e 9/6 + e ) It is not surprising that the second method yields the larger result, since the integrand e x2 is an increasing function, sof(x i ) >f(x i ). Figure 7.4. Left-endpoint method.

60 54 7. METHODS OF INTEGRATION Figure 7.5. Right-endpoint method. Furthermore, for each point x [x i,x i ], we have f(x i ) f(x) f(x i ). So the left-endpoint method underestimates the area of each strip S i, while the right-endpoint method overestimates it. Therefore, the area of D and hence the correct value of 0 ex2 dx is between the two values of.2759 and.7055 computed above. Replacing the value of n = 4 by some larger number will result in a more precise approximation (and more work). Using the midpoint method will result in an approximation A that is closer to the actual value of the integral 0 ex2 dx, but it is not completely obvious from which side A approximates 0 ex2 dx, that is, whether A< 0 ex2 dx or A> 0 ex2 dx More Advanced Approximation Methods Trapezoid Method. If the difference between f(x i ) and f(x i ) is large, then estimating the area of S i by using rectangles could lead to large errors. A more refined approach is to estimate the area of S i by computing the area of the trapezoid whose vertices are the points (x i, 0), (x i, 0), f(x i ), and f(x i ). We know that the area of this trapezoid is the average length of its parallel sides times the distance of those parallel sides from each other, that is, (f(x i)+f(x i ))(x i x i ) 2.

61 47. APPROXIMATE INTEGRATION 55 Figure 7.6. Trapezoid method. Summing over all possible values of i, we get an estimate for all the area of D, that is, for b f(x) dx. Indeed, we obtain the formula a b n (f(x i )+f(x i ))(x i x i ) f(x) dx =. a 2 i= In particular, if the x i are chosen so that they split the interval [a, b] into n equal parts, then the last displayed equation simplifies to b f(x) dx = b a n (f(x i )+f(x i )) a 2n i= = b a 2n (f(a)+2f(x )+2f(x 2 )+ +2f(x n )+f(b)). Note that f(x 0 )=f(a) and f(x n )=f(b) occur only once in the sum in the last line since a and b are each part of only one of the intervals [x i,x i ]. Example Use the trapezoid method with n =4to find the approximate value of 0 ex2 dx. Solution: We will have x =/4, x 2 =/2, and x 3 =3/4, just as in Example This yields e x2 dx = ( e 0 +2e /6 +2e /4 +2e 9/6 + e ) 0 8 =.4907.

62 56 7. METHODS OF INTEGRATION Figure 7.7. Trapezoid method. The alert reader may have noticed that the result we obtained is precisely the average of the left-endpoint and right-endpoint approximations we obtained for the same integral in the previous section. (This means that it is a better approximation than at least one of the two earlier ones.) This is not an accident, and in Exercise , the reader will be asked to prove that, under certain conditions, this phenomenon will always occur Simpson s Method. A similar method is Simpson s method, in which we use parabolas instead of straight lines for approximation. For simplicity, let us now assume that the points x i split the interval [a, b] into n equal parts, that is, x i x i =(b a)/n. In order to simplify the notation, let us set y i = f(x i ). For any integer i [,n ], consider the points (x i,y i ), (x i,y i ), and (x i+,y i+ ). There is exactly one parabola p i of the form y = Ax 2 + Bx + C that contains these three points. It can then be proved that the area under that parabola more precisely, the area of the domain P i bordered by the horizontal axis, the vertical lines x = x i and x = x i+ and p i is equal to b a (7.9) 3n (y i +4y i + y i+ ). If we summed (7.9) over all possible values of i, we would not get a good estimate, since most points of the domain under the curve would

63 47. APPROXIMATE INTEGRATION 57 be part of two of the P i. For instance, a point with a horizontal coordinate between x i and x i+ is part of both P i and P i+. Therefore, we sum the last displayed equation over all even values of i, and, accordingly, we stipulate that n be an even number. This leads to the following estimate. Theorem 7. (Simpson s Method). Let n be an even positive integer. Then b a f(x) dx b a 3n (y 0 +4y +2y 2 +4y y n 2 +4y n + y n ). Example Use Simpson s method with n =4to approximate sin 0 (x2 ) dx. Solution: We have y 0 = 0, y = sin(/6), y 2 = sin(/4), y 3 = sin(9/6), and y 4 = sin. So Simpson s method yields 0 sin ( x 2) dx (4 sin(/6) + 2 sin(/4) + 4 sin(9/6) + sin ) Figure 7.8. Simpson s method.

64 58 7. METHODS OF INTEGRATION Note that this result confirms our intuition in that if x [0, ], then sin (x 2 ) sin x, and so 0 sin ( x 2) dx 0 sin x = cos Bounds on the Error Term. The error term E of an approximation is the difference between the number obtained by the approximation and the actual value of the quantity that was approximated. (In this section, that actual value is the value of b f(x) dx.) It goes without a saying that the smaller the absolute value of the error term, the better the approximation is. The field of numerical analysis studies the error terms of approximation methods. The techniques of that field yield various bounds on error terms. We collected some of these bounds in the following theorem. Theorem 7.2. Let f be a twice-differentiable function on [a, b] such that f (x) M if x [a, b]. Then the following hold for the approximation methods used to compute b a f(x). (a) If E T is the error term of the trapezoid method, then M(b a)3 E T. 2n 2 (b) If E M is the error term of the midpoint method, then M(b a)3 E M. 24n 2 (c) If, for all x [a, b], the number f (4) (x) is defined and is at most as large as the constant K, and E S is the error term of Simpson s method, then K(b a)5 E S. 80n 4 Comparing the formulas of parts (a) and (b) of the previous theorem, we can conclude that the worst-case scenario of the midpoint method is better than the worst-case scenario of the trapezoid method. This, of course, does not mean that the midpoint method is always better than the trapezoid method. Example 7.3. Find an upper bound for the approximation obtained in Example 7.29.

65 48. IMPROPER INTEGRALS 59 Solution: We apply part (a) of Theorem 7.2. We have f(x) =e x2, so f (x) =e x2 (4x 2 + 2). This is an increasing function on [0, ], so its maximum is taken at x =, showing that f (x) 6e. As in Example 7.29, we chose n = 4, so the previous theorem yields E T M(b a)3 2n 2 = 6e Exercises. () Use n = 4 and the midpoint method to find the approximate value of 0 ex2 dx. (2) Use n = 4 and the trapezoid method to find the approximate value of 0 e x2 dx. (3) Use n = 4 and Simpson s method to find the approximate value of 0 sin x2 dx. (4) What value of n should we use in each of the preceding three exercises to get an error term that is less than 0 6? (5) Let us assume that the points x,x 2,...,x n used for approximate integration split the interval [a, b] inton equal segments. Prove that the result obtained by the trapezoid method will be the average of the left-endpoint method and the right-endpoint method. (6) How large is the error term of the approximation in exercise in the worst case? 48. Improper Integrals In our studies of integration, we have not dealt with definite integrals over infinite intervals, nor did we integrate functions over an interval if the function was not defined in every point of that interval. In this section, we will consider definite integrals of these kinds, which are called improper integrals Infinite Intervals. For finite intervals, we have identified b f(x) dx a with the area of the domain limited by the graph of f, the vertical lines x = a and x = b, and the horizontal axis. We learned that, by the fundamental theorem of calculus, the equality b f(x) dx = F (b) F (a) a holds, where F is an antiderivative of f. Now let us consider the integral of f over the infinite interval [a, ). Recalling that b f(x) dx is equal to a certain area, we intuitively want a f(x) dx to equal the area of the domain bordered by the line x = a, a

66 60 7. METHODS OF INTEGRATION Figure 7.9. Area under the curve y = f(x) from x = a to x = b. the horizontal axis, and the graph of f. Note that this area may be a finite number, even if it is not squeezed between two vertical lines. One example of this is when f(x) =0ifx>N for some real number N. Time has come to formally define f(x) dx. a Definition 7.. Let f be a function. If the integral b f(x) dx a b exists for all b>aand lim b f(x) dx = L exists as a (finite) real a number, then we say that the integral f(x) dx is convergent, and a we write f(x) dx = L. a b If lim b f(x) dx does not exist or is infinite, then we say that a f(x) dx is divergent. a Note that if F is an antiderivative of f, then lim b b a ( lim b F (b)) F (a). f(x) dx = lim b (F (b) F (a)) Therefore, the integral f(x) dx is convergent if and only if lim a b F (b) exists and is finite. Example Let f(x) =x 2. Compute f(x) dx.

67 48. IMPROPER INTEGRALS 6 Figure 7.0. Area under the curve y = f(x) fromto. Solution: We have x 2 dx = lim b b x 2 dx = lim b [ x ] b = lim b b + =. In particular, f(x) dx is convergent. Encouraged by the simple solution of the last example, we are going to compute the more general integral x r for any real number r. Example Let f(x) =x r. Compute f(x) dx. Solution: Let us first assume that r. Then we have x r dx = lim b b = lim b x r dx [ r + xr+ If r>, then r +> 0 and lim x x r+ =, so the limit in the last displayed row is infinite, and hence x r dx is divergent. If r<, then r +< 0 and lim x x r+ = 0, so the limit in the last displayed row is equal to, and hence x r dx is convergent. r+ ] b.

68 62 7. METHODS OF INTEGRATION If r =, then we need to compute x r dx differently, since, in that case, x a dx xa+. Instead, we have a+ x dx = lim b b x dx = lim b [ln x] b = lim b ln b =. Therefore, x dx is divergent. Note that the results of the previous example prove the following important theorem. Theorem 7.3. Let r be a real number. (i) If r, then x r dx is divergent. (ii) If r<, then x r dx is convergent. The following definition is not very surprising. It is the counterpart of Definition 7.. Definition 7.2. Let f be a function and let b be a real number such that, for all real numbers a<b, the integral b f(x) dx exists. If a L = lim a f(x) dx exists as a (finite) real number, then we say that the integral b f(x) dx is convergent, and we write b f(x) dx = L. If lim a f(x) dx is infinite or if it does not exist, then we say that b f(x) dx is divergent. The following definition makes it clear how and when we can define an integral on the entire line of real numbers. Definition 7.3. Let f be a function and let m be a real number such that both m f(x) dx and f(x) dx are convergent. Then we m say that the integral f(x) dx is convergent and that f(x) dx = m f(x) dx + Otherwise, we say that f(x) dx is divergent. See Figure 7. for an illustration. Example Compute e x dx. m f(x) dx.

69 48. IMPROPER INTEGRALS 63 Figure 7.. m f(x) dx is blue, while f(x) dx is orange. m Figure e x dx. Solution: We set m = 0 and apply Definition 7.3. We get that e x dx is convergent if both of 0 e x dx and e x dx are convergent. However, 0 0 e x dx = lim a [ e x ] 0 a =+ is divergent and therefore so is e x. Figure 7.2 shows the domain whose area is equal to 0 e x dx. The reader could ask how we knew that we needed to select 0, and not some other real number, for the role of m, that is, to split the real number line into two parts. The answer is that we did not, and other choices of m would have given the same result since the integrand

70 64 7. METHODS OF INTEGRATION converges to infinity as x goes to negative infinity. We chose m =0 because it was convenient to do so. Note that all improper integrals discussed in this section are called Type improper integrals Vertical Asymptotes. Sometimes we may want to compute the integral of a function f on a finite interval [a, b] so that in some point c [a, b], the function f has a vertical asymptote. An example is the function f(x) =/(x 2 4) on the interval [, 3]. In this case, we use the technique of limits to formally define b f(x) dx, as we did in the a previous section. Definition 7.4. Let f be a function that is continuous on [a, b], except for one point c [a, b]. Then we set (7.20) and c a b f(x) dx = lim t c t a f(x) dx (7.2) f(x) dx = lim f(x) dx. c t c + t Furthermore, if both of the two limits displayed above exist and are finite, we set b a f(x) dx = c a b f(x) dx + b c f(x) dx. Note that if the only point c in which f is not continuous is one of the endpoints of [a, b], then we only have to compute one of (7.20) and (7.2), since the other integral is taken over a trivial interval and is hence zero. Example Compute 0 x /2 dx. Solution: As the only point in [0, ] in which f(x) =x /2 is not continuous is 0, we use formula (7.2) with c = 0 and b =. We get 0 x /2 dx = lim x /2 dx t 0 + t = lim ] t 0 +[2x/2 t =2 lim t/2 t 0 + =2 0 =2.

71 48. IMPROPER INTEGRALS 65 Figure x /2 dx. So the integral 0 x /2 dx is convergent. Example Compute 4 x 2 dx. Solution: We apply Definition 7.4 since the interval [, 4] has one point, c = 0, where the integrand is not continuous. Therefore, 4 x 2 dx = 0 = lim t 0 = lim x 2 dx + t 4 0 x 2 dx x 2 dx + lim t t x 2 dx ] t t 0 [ x + lim ] 4 t 0 +[ x t = + =. So the integral in question is divergent. Figure 7.4 shows the domain whose area is equal to 4 0 x 2 dx and the correct way of breaking that interval up to two parts. Note that we would have reached the wrong conclusion if we had disregarded the fact that x 2 is not continuous at x = 0 and tried to apply the fundamental theorem of calculus. Indeed, in that case, we would have obtained the wrong result: [ x ] 4 = 4 = 5 4. This result is incorrect, and the incorrect step was to apply the fundamental theorem of calculus for a function that is not continuous in the entire interval of integration. The integrals that we have discussed in this section are called Type 2 improper integrals.

72 66 7. METHODS OF INTEGRATION Further Remarks. Figure x 2 dx Improper Integrals of Mixed Type. There are some integrals that are improper for two reasons. They are taken over an infinite interval, and that interval contains a point in which the function is not continuous. In that case, we split up the interval of integration so that now we have two integrals, one of which is of Type and the other of which is Type 2. Example Compute 0 dx. (x 2) 2 Solution: We break up the interval [0, ) to the union of the two intervals [0, 2] and [2, ), getting 0 (x 2) 2 dx = 3 0 (x 2) 2 dx + 3 (x 2) 2 dx. The first term on the right-hand side is an improper integral of Type 2, and the second term on the right-hand side is an improper integral of Type. We can compute both by the methods presented earlier in this section Comparison Test. Comparison tests for improper integrals work very similarly to those for proper integrals. Theorem 7.4. Let us assume that, for all x a, the chain of inequalities 0 f(x) g(x) holds. (i) If a (ii) if a f(x) dx is divergent, then so is a g(x) dx is convergent, then so is a Example Show that 3 g(x) dx. f(x) dx. dx is convergent. x 2 ln x

73 48. IMPROPER INTEGRALS 67 Figure dx. (x 2) 2 Figure dx. x 2 ln x Solution: If x 3, then ln x >, so x 2 ln x > x 2, and therefore the integrand is less than. See Figure 7.6 for an illustration. On x 2 the other hand, we know that /x 2 dx is convergent, so our claim 3 follows from the comparison test Exercises. () Is sin xdx convergent or divergent? (2) Is 0.5 x.5 dx convergent? (3) Is 3 x 2 dx convergent? (4) Is x 2 dx convergent? (5) Is xe x dx convergent? 0 (6) Is (7) Is 0 dx convergent? xe x2 dx convergent? e x +x

74

75 CHAPTER 8 Sequences and Series 49. Infinite Sequences A sequence can be thought of as an ordered list of numbers a,a 2,..., a n,a n+,... The subscript n indicates the position of a number a n in the sequence; for example, a is the first element, a n is the nth element, and so on. Definition 8. (Sequence). A sequence is a function f defined on the set of all positive integers; that is, it is a rule that assigns a number to each positive integer. If f(n) =a n for n =, 2,..., it is customary to denote the range of f by the symbol {a n } or {a n }. So a sequence can be defined by specifying the rule a n = f(n) to calculate the nth term from an integer n. For example, a n = n { n } { n + = n + 2, 2 3, 3 } 4,..., { a n = ( )n ( ) n } { =, n n 2, 3, } 4,..., a n = q n {q n } 0 = {, q,q 2,q 3,...}. Sequences can also be defined recursively, that is, by a relation that allows us to find a n if a m, m<n, are known. For example, the Fibonacci sequence {f n } is defined by the recurrence relation f = f 2 =,f n = f n +f n 2,n 3 {f n } = {,, 2, 3, 5, 8, 3,...} Graphic representation of sequences. A sequence can be pictured similarly to the graph of a function by plotting points (x, y) =(n, a n ), n =, 2,..., on the xy plane. For example, the sequence a n = n/(n+) is the set of points on the graph y = x/(x + ) corresponding to all positive integer values of x, that is, x =, 2, Limit of a Sequence. The sequence a n = n/(n + ) has the property that the values a n approach as n becomes larger. Indeed, the difference a n = n n + = n + 69

76 70 8. SEQUENCES AND SERIES Figure 8.. Set of points on the graph y = x/(x + ) corresponding to integer values x = n. For large x, x/(x+) approaches from below, and hence n/(n+) = /(+/n) asn. The difference n/(n+) = /(n + ) can be made smaller than any (small) number ε>0 for all n>n and some integer N. decreases with increasing n and hence can be made smaller than any preassigned positive number ε for all n>n, where N depends on ε. For example, put ε =0 2. Then the condition a n <εimplies that /(n +)<εor /ε <nor 99 <n, that is, a n < 0 2 for all n>99. If ε =0 4, then a n < 0 4 for all n>n= In other words, no matter how small ε is, there is only a finite number of elements of the sequence that lie outside the interval ( ε, +ε). In this case, the sequence is said to converge to the limit value. Definition 8.2 (Limit of a Sequence). A sequence {a n } has the limit a if, for every ε>0, there is a corresponding integer N such that a n a <εfor all n>n. In this case, the sequence is said to be convergent, and one writes lim a n = a or a n a as n. n If a sequence has no limit, it is called divergent. One can say that a sequence {a n } converges to a number a if and only if every open interval containing a has all but finitely many of the elements of {a n }. Theorem 8. (Uniqueness of the Limit). The limit of a convergent sequence is unique: lim n a n = a and lim n a n = a = a = a.

77 49. INFINITE SEQUENCES 7 Figure 8.2. Definition of the limit of a sequence. The dots indicate numerical values a n (vertical axis). The integer n increases from left to right (horizontal axis). The convergence of a n to a number a means that, for any small ε>0, there is an integer N such that all the numbers a n, n>n, lie in the interval (a ε, a + ε). It is clear that N depends on ε. Generally, a smaller ε requires a larger N. Proof. Fix ε>0. Then, by the definition of the limit, there are numbers N and N such that a n a ε if n>n and a n a ε if n>n. Hence, both inequalities hold for n>max(n,n ) and for all such n: 0 a a = a a n + a n a a n a + a n a < 2ε; that is, the nonnegative number a a is smaller than any preassigned positive number, which means that a a =0ora = a. Since a sequence is a function defined on all positive integers, there is a great deal of similarity between the asymptotic behavior of a function f(x) asx and a sequence a n = f(n). Theorem 8.2 (Limits of Sequences and Functions). Let f be a function on (0, ). Suppose that lim x f(x) =a. If a n = f(n), where n is an integer, then lim n a n = a. The validity of the theorem follows immediately from the definition of the limit lim x f(x) =a (i.e., given ε>0, there is a corresponding number M such that f(x) a <εfor all x>m) by noting that the range of f(x) contains the sequence a n = f(n). Example 8.. Find the limit of the sequence a n =lnn/n if it exists or show that the sequence is divergent.

78 72 8. SEQUENCES AND SERIES Solution: Consider the function f(x) =lnx/x such that a n = f(n) for all positive integers. Hence, ln n lim n n = lim ln x x x = lim /x x =0, where the indeterminate form arising from ln x/x as x has been resolved by means of l Hospital s rule. Note that l Hospital s rule applies not to sequences but to functions of a real variable. Following the analogy between the limits of sequences and functions, one can select a particular class of divergent sequences. Definition 8.3 (Infinite Limits). The limit lim n a n = means that, for every positive number M, there is a corresponding integer N such that a n >M for all n>n. Similarly, the limit lim n a n = means that, for every negative number M, there is a corresponding integer N such that a n <M for all n>n. Example 8.2. Analyze the convergence of the sequence a n =/n p, where p is real. Solution: Put f(x) =/x p for x>0. Then a n = f(n) and therefore lim n n p = lim x x p = 0 if p>0, if p =0, if p<0. Example 8.3. Analyze the convergence of the sequence a n = q n, n =0,,..., where q is real. Solution: Suppose q>0. Put f(x) =q x = e x ln q. From the properties of the exponential function, it follows that e ax if a =lnq>0, e ax =ifa =lnq = 0, and e ax 0ifa =lnq < 0. Therefore, a n if q>, a n = ifq =, and a n 0if0<q<. When q =0,a n = 0. Suppose q<0. Then q = q and a n =( ) n q n = ( ) n e n ln q. If q <, then even and odd terms of the sequence converge to 0: a 2n = e 2n ln q 0 and a 2n = e (2n ) ln q 0as n. When q =, the sequence a n =( ) n is divergent because a 2n = and a 2n = ; that is, the sequence oscillates between and for all n and a n does not approach any number. Finally, if q<, the sequence is divergent, too, because a 2n = e 2n ln q but a 2n = e (2n ) ln q. Moreover, it approaches neither nor as it oscillates taking ever-increasing positive and negative values.

79 49. INFINITE SEQUENCES 73 Thus, lim n qn = 0 if q (, ), if q =, if q>, and the sequence does not converge if q Subsequences. Given a sequence {a n }, consider a sequence {n k } of positive integers such that n <n 2 <n 3 <. Then the sequence {a nk }, k =, 2,..., is called a subsequence of {a n }. Recall that a sequence {a n } converges to a number a if and only if every open interval containing a has all but finitely many of the elements of {a n }. Therefore, {a n } converges to a if and only if every subsequence of {a n } converges to a. This necessary and sufficient criterion for convergence has already been used in Example 8.3. The sequence a n =( ) n does not converge because it has two subsequences a 2n = and a 2n =, which converge to different numbers, Limit Laws for Sequences. The limit laws for functions also hold for sequences, and their proofs are similar. If {a n } and {b n } converge to numbers a and b, respectively, and c is a constant, then lim n + b n ) = lim a n + lim b n = a + b, n n n lim n)=c lim a n = ca, n n lim nb n ) = lim a n lim b n = ab, n n n a n lim = lim n a n = a n b n lim n b n b if b 0, lim n) p = ( lim a n ) p = a p n n if p>0 and a n > 0. The squeeze theorem also applies to sequences. Theorem 8.3 (Squeeze Theorem). If c n a n b n for n>n and lim n b n = lim n c n = a, then lim n a n = a, where a can also be ±. Example 8.4. Find the limit of a n = sin(π/ n). Solution: Since x sin x x if x 0, one has c n = π/ n a n π/ n = b n, where c n 0 and b n 0asn. By the squeeze theorem, sin(π/ n) 0asn. Theorem 8.4. If lim n a n =0, then lim n a n =0.

80 74 8. SEQUENCES AND SERIES Figure 8.3. The squeeze theorem. The dots indicate numerical values (vertical axis) of the sequences b n (blue), c n (black), and a n (red). The integer n increases from left to right (horizontal axis). The sequences b n and c n converge to a number a. This means that the differences b n a and c n a can be made arbitrarily small for all n N and some integer N. Since c n a n b n, the difference a n a is also arbitrarily small for all n N. By the definition of the limit, the sequence a n must converge to a, too. This theorem follows directly from the definition of the limit of a sequence where a =0. Theorem 8.5. If a n a as n and the function f is continuous at a, then lim f(a n)=f(a). n This theorem asserts that if a continuous function is applied to the terms of a convergent sequence, the result is also convergent. Proof. The continuity of f at a means that lim x a f(x) =f(a) or, by the definition of this limit, for any ε>0, there is a corresponding δ>0 such that f(x) f(a) <εwhenever x a <δ. Having found such δ, put ε = δ and, by the definition of the limit lim n a n = a, for any such ε > 0, there is a corresponding integer N such that a n a <ε = δ if n>n. Therefore, for any ε>0, one can find a corresponding integer N such that f(a n ) f(a) <εfor all n>n, which means that lim n f(a n )=f(a).

81 49. INFINITE SEQUENCES 75 Example 8.5. Find the limit of the sequence a n = exp(/n 2 ). Solution: Consider the sequence b n =/n 2. Then lim b n = lim n x x =0. 2 Put f(x) =e x. Then a n = f(b n ). By continuity of the exponential function, lim a n = exp( lim b n )=e 0 =. n n Exercises. () Find a formula for the general term a n of the sequences: {a n } = {, 3, 5, 7, 9, },..., {a n } = {, 2, 4, 8, } 6,..., {a n } = { 4, 2 7, 3 0, 4 3, 5 } 6,.... In (2) (3), determine whether the sequence converges or diverges. If it converges, find the limit. (2) a n =2 n. (3) a n =2 n ( ) n 2 n. (4) a n =(3 5n 2 )/( + n 2 ). (5) a n = tan[nπ/(2+4n)]. (6) a n = sin 2 [π(n 2 +2)/(2n 2 + 5)]. (7) a n = ln(an)/ln(bn), where a and b are positive numbers. (8) a n = n p e n, where p is real. (9) a n = n cos(/n). (0) a n = (n 3 +)/(8n 3 +4n 2 +2n +). () a n = (ln n) p /n, where p>0. (2) a n = tan (n 2 ). (3) a n = n 2 + n n.

82 76 8. SEQUENCES AND SERIES 50. Special Sequences Theorem 8.6 (Special Sequences). Let p and q be real numbers. (8.) lim (8.2) lim (8.3) (8.4) (8.5) n n lim n lim n lim n n p = if p>0, n n =, n q =0 pn if p>, n! n =0, n q n n! =0. Proof. (8.). Ifp>, put a n = n p. Then a n > 0 and, by the binomial theorem, +na n ( + a n ) n = p. Note that all terms in (+a n ) n =+na n +n(n )a n /2+ +nan n +a n n are positive. So, by retaining only the first two terms, a smaller number is obtained, that is, + na n ( + a n ) n. It follows from this inequality that 0 <a n p n. By the squeeze theorem, a n 0asn and hence n p = a n +. The case p = is trivial. If 0 <p<, the result is obtained by taking reciprocals: lim n n p = lim n n (/p) = lim n n (/p) = because /p >. (8.2). Put a n = n n. Then a n 0 and, by the binomial theorem, n =(+a n ) n n(n ) a 2 2 n. Hence, for n 2, 2 0 a n n. By the squeeze theorem, a n 0or n n = a n + asn 0. (8.3). Consider the function f(x) =x q e cx, where c>0. By the asymptotic property of the exponential function, f(x) 0asx for any q; the exponential grows faster than any power function (which

83 50. SPECIAL SEQUENCES 77 has been proved in Calculus I). Since a n = f(n) for c =lnp>0if p>, one concludes that n q lim n p = lim n n nq e n ln p = lim x q e x ln p =0. x (8.4). The following inequality holds: a n = n! n = 2 3 n n n n n n = n 2 n 3 n n n 0 <a n n. By the squeeze theorem, a n 0asn. (8.5). Ifq>0, then there is a positive integer k such that k q<k, that is, k is the smallest positive integer such that q/k <. The following inequality holds: a n = qn n! = q q 2 q k q k q n qk q k q n q n qk q n = qk n. By the squeeze theorem, 0 <a n qk 0asn and a n n converges to 0. The case q = 0 is trivial. If q<0, then a n = q n /n! = q n /n! 0asn and hence a n converges to 0, too. Example 8.6. Find the limit of a n = n n q, where q>, if it exists or show that the sequence diverges. Solution: lim n by (8.2) and the basic limit laws. n nq = lim n ( n n) q = ( lim n n n) q = q = 50.. Monotonic Sequences. be Definition 8.4 (Monotonic Sequences). A sequence a n is said to for all n =, 2,... monotonically increasing if a n a n+, monotonically decreasing if a n a n+ The class of monotonic sequences consists of the increasing and the decreasing sequences. Example 8.7. Show that the sequence a n = n/(n 2 +) is monotonically decreasing.

84 78 8. SEQUENCES AND SERIES Figure 8.4. The sequence on the left is monotonically increasing, and the sequence on the right is monotonically decreasing. Solution: The inequality a n a n+ must be established. It is equivalent to the following inequalities obtained by cross-multiplication: n + (n +) 2 + n n 2 + (n + )(n2 +) n[(n +) 2 +] n 3 + n 2 + n + n 3 +2n 2 +2n n 2 + n. The latter inequality is true for n. Therefore, a n+ a n (in fact, the strict inequality a n+ < a n holds as well), and the sequence is monotonically decreasing. Definition 8.5 (Bounded Sequence). A sequence is said to be bounded above if there is a number M such that a n M for all n. A sequence is said to be bounded below if there is a number m such that m a n for all n. A sequence is said to be bounded if it is bounded above and below: m a n M for all n. For example, the sequence a n =/n is bounded: 0 <a n. The sequence a n = e n is bounded below, but not above. Completeness axiom for the set of real numbers. The completeness axiom for the set of real numbers says that if S is a nonempty set of real numbers that has an upper bound M (x M for all x S), then S has a least upper bound. By definition, the number a is a least upper upper bound of S if, for any ε>0, a ε is not an upper bound of S. The least upper bound is called the supremum of S and denoted sup S. Naturally, sup S M for any upper bound M of S. If S has a lower bound m, then it also has the greatest lower bound, denoted

85 50. SPECIAL SEQUENCES 79 Figure 8.5. A bounded sequence. The dots indicate numerical values of a n (vertical axis). The integer n increases from left to right (horizontal axis). All the numbers a n lie in the interval: m a n M. inf S (the infimum of S). The number inf S is a lower bound of S such that inf S + ε is not a lower bound of S for any positive ε>0; that is, m inf S for any lower bound of S. The completeness axiom is an expression of the fact that there is no gap or hole in the real number line. Theorem 8.7 (Monotonic Sequence Theorem). Suppose {a n } is monotonic. Then {a n } converges if and only if it is bounded. Proof. Suppose a n a n+ (the proof is analogous in the other case). Let S be the range of {a n }. If {a n } is bounded, let a be the least upper bound of S (it exists by the completeness axiom). Then a n a for all n. For every ε>0, there is an integer N such that a ε<a N a; otherwise, a ε would be an upper bound of S. Since {a n } increases, the inequality n N implies that a ε<a n a a a n <ε whenever n N, which shows that {a n } converges to s. Example 8.8. Investigate the convergence of the sequence defined by the recurrence relation a =2and a n+ = 2 (a n +3). Solution: Let us compute the first few terms of the sequence a =2, a 2 =2.5, a 3 =2.75, and so on. The initial terms suggest that the sequence is monotonically increasing, and one can try to prove this property a n a n+ for all n. A commonly used technique to do so

86 80 8. SEQUENCES AND SERIES Figure 8.6. Monotonic sequence theorem. A bounded monotonic sequence with numerical values indicated by dots (vertical axis). The integer n increases from left to right. If S = sup n {a n } is the least upper bound of all a n, then, for any number ε>0, S ε is not an upper bound of the sequence. Since a n increases monotonically, there is an integer N such that all the numbers a n, n>n, are greater than S ε and hence lie in the interval (S ε, S). This means that a n converges to S. is mathematical induction. The statement is true for n =. Suppose that the statement is true for n = k, then one has to prove that the statement is also true for n = k +. If the proof goes through, then starting with n =, one can establish the statement for n =2,n =3, an so on. This is the basic idea of mathematical induction. Using the recurrence relation, a k+ >a k = 2 (a k+ +3)> 2 (a k +3) = a k+2 >a k+. Thus, the sequence is indeed monotonically increasing. If it happens to be bounded, then it converges. Again, mathematical induction turns out to be helpful. The first terms suggest that a n < 3. This is true for n =. Suppose the inequality is true for n = k. Let us try to prove that this assumption implies that the inequality holds for n = k +. Using the recurrence relation, a k < 3 = 2 (a k +3)< 2 (3+3) = a k+ < 3. Thus, the sequence is monotonic and bounded and hence converges. If the sequence a n converges to a, then so does the sequence a n+k for any integer k (in the definition of the limit, change N to N + k to prove this). Since the existence of the limit has been established, one can

87 50. SPECIAL SEQUENCES 8 take the limit of both sides of the recurrence relation lim n+ = n 2 ( lim n +3) n = a = (a +3) 2 = a =3. Thus, a n 3asn. Example 8.9. Investigate the convergence of the sequence defined by the recurrence relation a = 2 and a n+ = 2+a n. Solution: The first few terms of the sequence suggest that the sequence is increasing: a = 2 <a 2 = Let us try to prove the inequality a n <a n+ by induction. Suppose it is true for n = k. Then, by monotonicity of the square root function and the recurrence relation, a k <a k+ a k < a k+ 2+a k < 2+a k+ a k+ <a k+2. The first terms of the sequence suggest that a < 3 and a 2 < 3. Let us try to prove that a n < 3 for all n by induction. Suppose the inequality holds for n = k. Then, by the recurrence relation, a k < 3 = 2+a k < 5 = 2+a k < 5 < 3 = a k+ < 3. Thus, the sequence is monotonic and bounded, and hence it converges. If its limit is a, then lim a n+ = lim 2+an = 2 + lim a n a = 2+a a =2. n n n Exercises. In () (5), find the limit of the sequence {a n } or show that it does not exist. () a n = n 2n (2) a n = cos 2n (n 2 )/n. (3) a n = nr n, where r is real. (4) a n =(2n )!!/(2n) n, where (2n )!! = 3 5 (2n ). (5) a n = n 3 n +5 n. In (6) (0), determine whether the sequence is monotonic or not monotonic. Is the sequence bounded? (6) a n =( 2) n. (7) a n =( ) n n. (8) a n = ne n. (9) a n = n + n.

88 82 8. SEQUENCES AND SERIES (0) a n = sin(qn)/n, where q is real. In () (5), find the limit of the sequence or show that it does not exist. () a = and a n+ =4 a n. (2) a = and a n+ =/( + a n ). (3) a = and a n+ =3 /a n. (4) a = 2 and a n+ =/(3 a n ). (5) a = and a n+ =+/( + a n ). (6) The size of an undisturbed fish population has been modeled by the formula p n+ = bp n /(a + p n ), where p n is the fish population after n years and a and b are positive constants that depend on the species and the environment. Suppose that p 0 > 0. Show that p n+ < (b/a)p n. Then prove that p n 0 if a>b; that is, the population dies out. Finally, show that p n b a if b>a. Hint: Show that p n is increasing and bounded, 0 <p n <b a if p 0 <b a. Ifp 0 >b a, then p n is decreasing and bounded, p n >b a. 5. Series 5.. Basic Definitions and Notation. With a sequence {a n }, one can associate a sequence {s n }, where n s n = a k = a + a a n. The symbol (8.6) k= a n = a + a 2 + a 3 + n= is called an infinite series, or just a series. The numbers s n are called the partial sums of the series (8.6). The limits of summation are often omitted to denote a series; that is, the symbol a n also stands for an infinite series. If {s n } converges to s, then the series is said to converge and one writes n= a n = s or lim n n a k = s. The number s is called the sum of the series. If the sequence of partial sums {s n } diverges, the series is said to diverge. It should be understood that s is the limit of a sequence of sums, and it is not obtained merely by addition. k=

89 5. SERIES 83 For example, the sequence of partial sums for the series ( ) n is s =, s 2 = + = 0, s 3 = s 2 =, or, generally, s n =(( ) n )/2. This sequence diverges as it has two subsequences s 2n = 0 and s 2n =, which converge to different numbers, 0. If one simply uses addition, different values for the sum of the series may be obtained: a n =(a + a 2 )+(a 3 + a 4 )+(a 5 + a 6 )+ =0+0+ =0, n= a n =(a + a 2 + a 3 )+(a 4 + a 5 + a 6 )+ = =, n= a n =a +(a 2 + a 3 )+(a 4 + a 5 )+ = =. n= Generally, by grouping terms in the sum in different ways (according to the associativity of addition), the sum is found to be any integer! The reader is advised to verify this. Thus, the addition rules cannot generally be applied to evaluate the sum of a series Geometric Series. Take a piece of rope of length m. Cut it in half. Keep one half and cut the other half in two pieces of equal length. Keep doing this, that is, keeping one half and cutting the other half in two equal-length pieces. The total length of the retained pieces is = ( ) 4 + = 2 n=0 2 n. This series must converge. The partial sum s n here is the total length of retained pieces. The sequence {s n } is monotonically increasing (after each cut piece of rope is added) and bounded by the total length. So it converges. From the geometry, it is also clear that s n =/2 n, where n is the number of cuts, and hence s n as one would expect (the total length of the rope). So it is concluded 2 =2. n n=0 This series is an example of the geometric series: +q + q 2 + q 3 + = q n, n=0

90 84 8. SEQUENCES AND SERIES where q is a number. The geometric series does not converge for any value of q. Theorem 8.8 (Convergence of a Geometric Series). A geometric series n= qn converges if q <, and, in this case, q n =, q <, q n=0 and the series diverges otherwise. Proof. If q =, the sequence of partial sums obviously diverges. If q, one has s n =+q + q q n = qs n = q + q 2 + q q n. Subtracting these equations, one infers Therefore, s n qs n = q n = s n = qn q. lim s q n n = lim n n q = q q lim n qn. It has been found that the sequence a n = q n converges only if q <, and, in this case, q n 0asn. If q, the geometric series diverges. Example 8.0. Analyze the convergence of the series Solution: The series can be written in the form 4q 0 +4q +4q 2 +4q 3 +, where q = 2/3. So its partial sums are four times the partial sum of the geometric series with q = 2/3. Therefore, ( 4 2 ) n ( =4 2 ) n 4 = 3 3 ( 2) = 2 5. n= n= 3 When real numbers are presented in decimal from, one often encounters a situation when a number has a repeated pattern of decimal places. Take, for example, the number ; that is, the combination 3 repeats itself in all decimal places starting in the second decimal place. Example 8.. Is the number rational or irrational? If it is rational, write it as a ratio of integers.

91 5. SERIES 85 Solution: By definition of the decimal representation, = = n=0 ( 00 = = = ) n 5.3. Necessary Condition for a Series to Converge. The following theorem follows from the limit laws applied to the sequences of partial sums. Theorem 8.9 (Properties of Series). Suppose that the series a n and b n are convergent and their sums are s and t, respectively. Let c be a number. Then the series (a n + b n ) and ca n converge and (an + b n )= a n + b n = s + t, (can )=c a n = cs. Indeed, if {s n } and {t n } are the sequences of partial sums of the series a n and b n, respectively, then the partial sums of the series (an + b n ) and (ca n ) are s n + t n and cs n, respectively. By the limit laws, s n + t n s + t and cs n cs. Note that the convergence of the series (a n + b n )doesnot imply the convergence of a n and b n. For example, put a n = and b n =. The series (a n + b n )= 0 = 0, while a n = and bn = ( ) diverge, and the equality (a n + b n )= a n + b n becomes meaningless ( 0 = ). This shows that the rules of algebra for finite sums are not generally applicable to series. Only series from a special class of absolutely convergent series, discussed later, behave pretty much as finite sums. It is clear that every theorem about sequences can be stated in terms of series by putting a = s and a n = s n s n for n>and vice versa. In particular, if the series converges, that is, s n s as n, then one can take the limit on both sides of this recurrence relation and conclude that lim n a n = lim n (s n s n )=s s = 0; that is, for a convergent series a n, the sequence {a n } necessarily converges to 0. Theorem 8.0 (Necessary Condition for a Series to Converge). If the series a n converges, then lim n a n =0. The converse is not generally true; that is, the condition lim n a n = 0 is not sufficient for a series to converge. However, it can still be used as a test for divergence of a series.

92 86 8. SEQUENCES AND SERIES Corollary 8. (Test for Divergence of a Series). If the limit lim n a n does not exist or if lim n a n 0, then the series a n diverges. Example 8.2. Show that the series n 3 /(3n 3 +) diverges. Solution: lim a n 3 n = lim n n 3n 3 + = lim n 3+ = 3 0, n 3 so the series diverges. If the necessary condition is satisfied, the series may converge or diverge. The sequence of partial sums has to be analyzed. Example 8.3. Find the sum of the series n= or show that it does not exist. n(n+) if it exists Solution: The necessary condition for convergence is evidently satisfied. So the sequence of partial sums has to be analyzed for convergence: n n ( s n = k(k +) = k ) k + = k= ( 2 k= ) ( + 2 3) ( + 3 4) + + = as n. n + So the sequence {s n } converges to and hence n= n(n +) =. ( n ) n + This example is a particular case of a telescopic series. Theorem 8. (Convergence of a Telescopic Series). A telescopic series n= (a n a n+ ) converges if lim n a n = a, and, in this case, (a n a n+ )=a a, n= The proof is analogous to the above example and based on the fact that the sequence of partial sums of a telescopic series s n = a a n+ converges to a a. The details are left to the reader as an exercise.

93 52. SERIES OF NONNEGATIVE TERMS Exercises. In () (3), determine whether the geometric series converges or diverges. () n= n=0 π n 3 n+ (2) n= n= e n 3 n (3) n=0 ( 5) n 3 2n In (4) (9), determine whether the series converges or diverges. If it converges, find its sum. Here p is a positive number, p>0. k n (4) (5) (6) n p k 2 + k + 5 n n= n= n= (7) (sin p) n e n ( (8) (9) e 2n 4 ) + n p n(n +) In (0) (2), determine whether the series converges or diverges by expressing it as a telescopic series. Find the sum of the series if it exists. (0) n=2 n 2 () n= 3 n 2 +3n +3 n= (2) n ln n + In (3) and (4), express the number as a ratio of integers (3) (4) In (5) (7), find the values of x for which the series converges. Find the sum of the series for those values of x. (5) n= x n 2 n (6) n= sin n x 3 n (7) In (8) and (9), solve the equation. (8) ( + x) n = 3 (9) n=2 n= (x 5) n n= e nx =9 n=0 52. Series of Nonnegative Terms In many applications, the terms of a series decrease monotonically. It appears that there is a relation between convergence of such series and convergence of improper integrals over an interval [, ). This relation allows one to establish a necessary and sufficient condition for series of nonnegative terms to converge.

94 88 8. SEQUENCES AND SERIES 52.. The Integral Test. Suppose f(x) is a positive, continuous, monotonically decreasing function on [, ) such that f(x) 0. Suppose also that the improper integral f(x) dx = lim a a 0 f(x) dx = I f exists. The value I f is the area under the graph y = f(x) over the interval [, ). Consider the series n= f(n). The necessary condition for convergence is fulfilled as f(n) 0asn. To investigate the convergence of the series, one has to analyze the convergence of its partial sums: n s n = f(k) =f() + f(2) + + f(n). k= Since the function f(x) monotonically decreases and is continuous on every interval [k, k + ], it attains its minimal and maximal values f(k +) f(x) f(k) on this interval and therefore (8.7) f(k +) k+ k f(x) dx f(k). This inequality leads to the following upper and lower estimates of the partial sums: s n f() + s n so that (8.8) 2 2 f(x) dx + n+ f(x) dx n n f(x) dx + + f(x) dx s n f() + f(x) dx = f() + n+ n n f(x) dx = n n+ f(x) dx, f(x) dx, f(x) for all n. This inequality shows that the following theorem holds. Figure 8.7. Integral test. An illustration of inequality (8.7).

95 52. SERIES OF NONNEGATIVE TERMS 89 Theorem 8.2 (Integral Test). Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f(n). Then the series n= a n converges if and only if the improper integral f(x) dx converges. In other words, f(x) dx converges = f(x) dx diverges = f(n) n= f(n) n= converges, diverges. Proof. If the improper integral converges to a number I f, then by (8.8) the sequence of partial sums is bounded, s n f() + I f, and monotonically increases, s n s n + f(n +)=s n+. Therefore, it is convergent. If the improper integral diverges, then, for any number M > 0, there is an integer N such that n+ f(x) dx M for all n>n. By the left inequality of (8.8), M s n for all n>n; that is, {s n } is a monotonically increasing, unbounded sequence and hence it diverges. Remark. Suppose that a n = f(n), where f(x) is a function on [, ), such that it is continuous, positive, and decreasing on [N, ), where N is an integer. Then (8.9) a n converges f(x) dx converges; n= that is, the integral test applies even if the sequence a n becomes monotonically decreasing only for n N. This is easy to understand by isolating the first N terms in the series a n = a + a a N + n= N a n = a + a a N + n=n b n, where b n = a N+n. Convergence of b n implies convergence of a n and vice versa as they differ by a number. Put b n = g(n), where g(x) = f(x+n ), which is a continuous, positive, decreasing function on [, ), and g(x) dx = f(x + N ) dx = by changing the integration variable u = x + N. N f(u) du n=

96 90 8. SEQUENCES AND SERIES Special Series of Nonnegative Terms. Theorem 8.3. The p-series n p n= converges if p> and diverges if p. Proof. If p 0, the series diverges because the necessary condition for convergence is not fulfilled, a n if p<0 and a n = 0if p =0. Forp>0, consider the function f(x) =x p, which is positive, continuous, and decreasing on [, ), and { ( ) a dx x = if p, p p a p ln a if p =. So, by the integral test, the series converges if p > because the improper integral diverges if 0 <p and converges if p> (the limit a exists only if p>). Note that the series n p diverges for all 0 <p despite that the necessary condition to converge is fulfilled: a n = n p 0. In particular, the harmonic series n= diverges. n The sum of a p-series ζ(p) = n p depends on the value of p>; that is, this series defines a function on (, ). This function is called Riemann s zeta function. Example 8.4. Investigate the convergence of the series n= (n +2) 3/2. Solution: The series can be written as (n +2) = 3/2 n= n=3 = n3/ /2 n= n 3/2. The latter series is a p-series that converges for p =3/2 >. Theorem 8.4. The series n(ln n) p n=2 converges if p>, and it diverges if p. Proof. Consider the function g(x) =x(ln x) p for x>. Its derivative reads g (x) = (ln x) p (p +lnx). If p 0, then g (x) > 0 for all x > and g(x) increases, while its reciprocal f(x) = /g(x) should

97 52. SERIES OF NONNEGATIVE TERMS 9 decrease. If p<0, then g (x) > 0 for all x>e p and hence g(x) increases, while f(x) =/g(x) decreases if x>e p >. Thus, for any p, there is an integer N such that the function f(x) =/[x(ln x) p ] is continuous, positive, and decreases on [N, ). By the integral test (8.9), the series in question converges if and only if the improper integral N dx x(ln x) p = converges, where the integration variable has been changed, u = lnx, du = dx/x. This integral diverges if p and converges if p>, and the conclusion of the theorem follows Estimate of the Sum. If a partial sum s n is used to estimate the sum of a convergent series of nonnegative terms f(n), how good is such an estimate? The remainder s s n has to be investigated to answer this question. Corollary 8.2 (Estimate of Sums). Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f(n). If the series an converges to a number s, then ln N du u p n+ f(x) dx s s n where {s n } is the sequence of partial sums. n f(x) dx, Proof. The first inequality is obtained by taking the limit n in (8.8) with the result (8.0) f(x) dx a n f() + f(x) dx, n= which is a legitimate operation because (8.8) holds for all n and the series converges (and so does the improper integral by the integral test). The remainder estimate is obtained by subtracting (8.8) from (8.0). Note the value of the improper integral does not coincide with the sum; it only determines an interval (8.0) in which the sum of a series lies. Example 8.5. Test the series n= (n2 +) for convergence or divergence. If it converges, estimate its sum. Solution: Put f(x) =(x 2 +), which is a continuous, positive, decreasing function on [, ), such that the series in question is f(n).

98 92 8. SEQUENCES AND SERIES Therefore, the integral test applies, and the series converges because dx x 2 + = lim a tan x a = lim tan a π a 4 = π 2 π 4 = π 4. By (8.0), its sum lies in the interval π 4 s f() + π 4 = 2 + π 4. Example 8.6. Test the series n= ne n for convergence or divergence. If it converges, estimate its sum. Solution: Consider the function f(x) =xe x. Since f (x) =e x xe x =( x)e x 0ifx, the function decreases on [, ), and the integral test applies to assess convergence of the series f(n): xe x dx = xde x = lim xe x a + a e x dx = e + e = 2 e, where the integration by parts has been used to evaluate the integral. The series converges to a number s that lies in the interval 2e s f()+2e =3e. Example 8.7. Estimate values of Riemann s zeta function ζ(p). How many terms does one need to retain in the partial sum s n to approximate ζ(p) correct to N decimal places? Solution: Riemann s zeta function is defined by the sum of the series ζ(p) = n= n p.forp>, dx x = lim x p a a p lim p a p = a p p = p. Since f() =, by (8.0), By Corollary 8.2, p ζ(p) 0 ζ(p +) s n n p p. dx x p+ = pn p. If ζ(p + ), p>0, is to be approximated by s n correct to N decimal places, then the remainder should be less than 5 0 N, which yields the condition on the number of terms: pn p < 5 0 N or n p > 0 N+ /(5p) orn p 0 N+ /(5p).

99 53. COMPARISON TESTS Exercises. In () (9), determine whether the series converges or diverges. (ln n) 4 n ln n () (2) (3) n 9/8 n n 2 n= n=2 n=2 n 2 2n 5 n (4) (5) (6) 3n 7/3 n 2 4n +5 n 4 + n= n= n= e /n 2n + tan n (7) (8) (9) n 2 n(n +) n 2 + n= n= In (0) (4), determine the values of p for which the series is convergent. (0) () n( + n 2 ) p (2) n p e n n ln n(ln(ln n)) p n=3 n= n= ( p (3) p ln n, p > 0 (4) n ) n + n= n= (5) How many terms of the series in Theorem 8.4 would one need to add to find its sum correct to N decimal places? (6) Show that the sequence n= a n = n ln n converges. The limit lim n a n = γ is called the Euler number. Hints: () Use (8.8) to show that if s n is the partial sum of the harmonic series, then s n +lnn and hence a n (i.e., the sequence {a n } is bounded). (2) Interpret a n a n+ as a difference of areas to show that {a n } is monotonic. 53. Comparison Tests Consider the series a n = n 4 +. n= n= This series has terms smaller than the corresponding terms of the convergent p-series: b n = n 4 n= n=

100 94 8. SEQUENCES AND SERIES because a n <b n for all n. If s n is the partial sum for a n and t n is the partial sum for b n, then s n <t n. Since t n converges to a number t, it is bounded, t n <t, and hence s n <t; that is, the sequence {s n } is monotonic and bounded, and therefore it converges. This line of arguments admits a generalization. Theorem 8.5 (Comparison Test). Suppose that a n and b n are series such that a n 0 and b n 0 for all n N and some integer N. Then bn converges and a n b n for all n N = a n converges, bn diverges and a n b n for all n N = a n diverges. Proof. The series n= a n and n=n a n differ by a number a + a a N. So convergence of n=n a n implies convergence of n= a n and vice versa. Therefore, it is sufficient to consider the case N =. The sequences of partial sums {s n }, s n = a + a a n, and {t n }, t n = b +b 2 + +b n, are monotonically increasing sequences because a n 0 and b n 0. If b n converges, then t n t as n and t n t for all n. By the hypothesis a n b n for all n, and therefore s n t n t, which shows that {s n } is monotonic and bounded and, hence, converges. If b n diverges, then t n. From the hypothesis a n b n, it follows that s n t n. Thus, s n as n. When applying the comparison test, the convergence properties of the series b n must be known. In many instances, a good choice is a geometric series (Theorem 8.8), a p-series (Theorem 8.3), a telescopic series (Theorem 8.), and the series in Theorem 8.4. Example 8.8. Test the series n= (2n +)/(3n3 + n 2 +) for convergence. Solution: Since a n is a rational function of n, a convenient choice of a series in the comparison test is a p-series: The series 2n + 3n 3 + n 2 + n= 2n + 3n 3 = 2 3 < 2n + 3n 3 = 2 3 n= n n n= converges as the sum of two convergent p-series. n 3. Example 8.9. Test the series n= ( n + n) for convergence. n 3

101 53. COMPARISON TESTS 95 Solution: One has a n = n + n = ( n + n)( n ++ n) n ++ n = n ++ n 2n + n = + = b n. 2 n The p-series / n diverges and so does the series in question by the comparison test. Theorem 8.6 (Limit Comparison Test). Suppose that a n and bn are series with positive terms. Let c = lim n (a n /b n ). If c =0and b n converges, then a n converges. If 0 <c<, then a n converges if only if b n converges. If c = and b n diverges, then a n diverges. Proof. If c = 0, then there is an integer N such that a n /b n < for all n>nby the definition of the limit. Hence, a n <b n for all n>n. If b n converges, then a n converges by the comparison test. If c (0, ), then, by the definition of the limit, for any number c>ε>0, there is an integer N such that c a n a n <ε m = c ε< <c+ ε = M for all n>n. b n b n Therefore, mb n <a n <Mb n for all n>n. By the comparison test, convergence of b n implies convergence of an due to the inequality a n <Mb n. The divergence of b n implies divergence of a n, again by the comparison test as mb n <a n. If c =, then, for any M>0, there is an integer N such that a n /b n >M when n>n. The inequality a n >Mb n shows that divergence of b n implies divergence of a n by the comparison test. It is often helpful to investigate the asymptotic behavior of a n as n to identify a suitable b n in the limit comparison test. Example Test the series n= (2n3/2 + n)/ n 6 + n 4 +for convergence. Solution: Let us find the asymptotic behavior of a n as n. For large n, the top of the ratio behaves as 2n 3/2, while the bottom of the ratio behaves as (n 6 ) /2 = n 3. Therefore, a n = 2n3/2 + n n6 + n 4 + = 2n3/2 ( 2n3/2 =2n 3/2 = b n n 3 n n 2 n 6 2 n )

102 96 8. SEQUENCES AND SERIES in the asymptotic region n. This shows that the ratio a n /b n converges to c =asn. By the limit comparison test, the series an converges because the p-series b n =2 n 3/2 converges. Example 8.2. Test the series n= (n5 +3 n )/ n 3 +5 n for convergence. Solution: Recall that the power function increases more slowly than the exponential function, that is, n p q n 0asn for any q> and any p. Hence, the asymptotic behavior of a n is a n = 3n ( + n 5 3 n ) 5 n ( + n 3 5 n ) 3n 5 n =(3/5)n = b n. This shows that the ratio a n /b n converges to c =asn. By the limit comparison test, a n converges because the geometric series bn converges (q =3/5 < ) Estimating Sums. If a series a n converges by comparison with a series b n, then the sum of a n can be estimated by comparing remainders for the series b n. Indeed, put a n = s and b n = t. Let {t n } and {s n } be the sequences of partial sums for b n and a n, respectively. The remainders satisfy the inequality: s s n = a n+ + a n+2 + b n+ + b n+2 + = t t n. So the accuracy of the approximation s s n is the same or higher than that of the approximation t t n. If, for example, one finds that n = N is sufficient for the equality t = t N to be correct to a specific number of decimal places, then s = s N is also correct to that or even a higher number of decimal places. The remainder is easy to estimate when b n = f(n), where the function f is simple to integrate, t t n f(x) dx. n Example Determine how many terms are needed to estimate the sum of the series n= tan (n 2 )/(n 3 +) correct to five decimal places. Solution: The function tan x is monotonically increasing for x>0 approaching asymptotically the value π/2. Therefore, Hence, s s n t t n π 2 a n = tan (n 2 ) n 3 + n dx x 3 = π 2 n 3 + π 2 n 3 = b n. π 4n 2 < n> π

103 54. ALTERNATING SERIES Exercises. In () (2), determine whether the series converges or diverges. () (4) (7) (0) n= n= n= n= n n 5/3 + n /3 + +( ) n n 3/2 + e / n n + (8) n +/n () (5) (2) n= n=2 n(ln n) 4 n n3 + n + ) ( sin 2 (9) n n= n= ( n n ) n (2) n= (3) n! n n (6) n= n=2 n= n 2 n! cos 2 (n) n 2 +2 n n 2 +2 n (3) How many terms does one need in the partial sum to estimate the sum of the series n= sin3 n/(n 3 + n) up to five decimal places? (4) Consider a sequence {a n }, where a n can take any value from the set {0,, 2,..., p }, where p> is an integer. The meaning of the representation of a number 0.a a 3 a 3... with base p is that (8.) 0.a a 3 a 3... = a p + a 2 p + a 3 2 p +. 3 When p = 0, the decimal system is obtained. The binary representation corresponds to p = 2. The Maya used p = 20 (the number of fingers and toes). The Babylonians used p = 60. Show that the series (8.) always converges. (5) Show that if a n > 0 and a n converges, then ln( + a n ) converges, too. (6) Prove that the convergence of a n, where a n > 0, implies the convergence of an /n. (7) If a n converges and if the sequence {b n } is monotonic and bounded, prove that a n b n converges. 54. Alternating Series Definition 8.6 (Alternating Series). Let {b n } be a sequence of nonnegative terms. The series ( ) n b n = b b 2 + b 3 b 4 + b 5 is called an alternating series.

104 98 8. SEQUENCES AND SERIES For example, the series (8.2) = ( ) n is an alternating series. It is called the alternating harmonic series. Theorem 8.7 (Alternating Series Test). If a sequence of positive terms {b n } is monotonically decreasing and lim n b n =0, then the alternating series ( ) n b n converges: (i) b n+ b n for all n (ii) lim n b n =0 = n= n ( ) n b n n= converges. Proof. The convergence of the sequence of partial sums {s n } is to be established. Consider a subsequence of even partial sums {s 2k }. One has s 2 = b b 2 0, s 4 = s 2 +(b 3 b 4 ) s 2, and, in general, s 2k = s 2(k ) +(b 2k b 2k ) s 2(k ) s 2(k 2) s 2 0 by the monotonicity of the sequence {b n }. Thus, the subsequence {s 2k } is monotonically increasing. By regrouping the terms in a different way, one can see that s 2k = b (b 2 b 3 ) (b 4 b 5 ) (b 2k 2 b 2k ) b 2k b because all numbers in parentheses are nonnegative by hypothesis (i), which shows that {s 2k } is also bounded. Therefore, it converges by the monotonic sequence theorem: lim s 2k = s. k For the subsequence of odd partial sums s 2k+ = s 2k + b 2k+, one infers by the limit laws and hypothesis (ii) that lim s 2k+ = lim s 2k + lim b 2k+ = s +0=s. k k k The convergence of two particular subsequences of a sequence to the same number s does not generally guarantee that the sequence converges to s (all its subsequences should converge to s). By definition, the limits of {s 2k } and {s 2k+ } mean that, given any number ε>0, there are positive integers N and N 2 such that s 2k s <εif k>n and s 2k+ s <εif k>n 2. Put N = max(2n, 2N 2 + ). Then s n s <εfor all n>n, which means that s n s as n. By this test, the alternating harmonic series (8.2) converges because the sequence b n =/n is monotonically decreasing and converges to 0.

105 54. ALTERNATING SERIES 99 Figure 8.8. Alternating series test. An illustration of its proof where two subsequences, s 2k and s 2k, of the sequence s n of partial sums are analyzed for convergence. Example Test the series n= sin(πn/2)/n for convergence. Solution: One has sin(πn/2)=, 0,, 0,,... for n =, 2, 3, 4, 5,..., respectively, or, in general, for odd n =2k, sin(πn/2)=( ) k, while for even n =2k, sin(πn/2) = sin(πk) = 0. Thus, the series in question is an alternating series: n= sin(πn/2) n = ( ) n 2n = ( ) n b n, b n = n= n= 2n. The sequence {b n } is monotonically decreasing and b n 0asn. So the series converges by the alternating series test. Remark. It should be noted that Theorem 8.7 provides only a sufficient condition for an alternating series to converge. So there are convergent alternating series that do not satisfy the hypotheses of Theorem 8.7. For example, the alternating series with b n = sin 2 (πn/q)/n 2, where q is an integer, is convergent by the comparison test because ( ) n+ b n = b n /n 2 and the p-series /n 2 converges (see the next section on absolutely conversing series). However, the sequence {b n } is not monotonically decreasing because b n 0 and it has a zero

106 00 8. SEQUENCES AND SERIES subsequence b kq =0,k =, 2,... So b n oscillates between the zero sequence and the sequence /n 2. Remark. Hypothesis (i) of Theorem 8.7 may be weakened (i) b n+ b n for all n N for some integer N. Indeed, ( ) n b n = b b 2 + b 3 b N + ( ) n b n n= n=n = b b 2 + b 3 b N +( ) N ( ) n c n, where c n = b n+n. The series ( ) n b n and ± ( ) n c n differ by a number, and therefore the convergence of one of them implies the convergence of the other. The series ( ) n c n converges by Theorem 8.7 as c n+ c n for all n and lim n c n = lim n b n+n = 0. Example Test the series n= ( )n n p /(n+) for convergence if p<. Solution: Here b n = n p /(n + ) and, for p<, lim b n p n = lim n n n + = lim n p n + = lim n p =0. n n So hypothesis (ii) of Theorem 8.7 is fulfilled. However, the monotonicity of {b n } is not obvious. To investigate it, consider the function f(x) =x p /(x + ), where x. If f(x) monotonically decreases, then so does the sequence b n = f(n). The condition f (x) 0 has to be verified: f (x) = pxp (x +) x p 0 (p )x p + px p 0 (x +) 2 p x( p) If p 0, this is true as x. If 0 <p<, then f(x) monotonically decreases for x p/( p) and one can always find an integer N p/( p) such that b n+ <b n for all n>n. So the series converges for all p< Estimating Sums of Alternating Series. A partial sum s n of any convergent alternating series can be used as an approximation of the total sum s, but this is not of much use unless the accuracy of the n=

107 54. ALTERNATING SERIES 0 approximation is assessed. The following theorem asserts that the absolute error of the approximation s s n does not exceed the value of b n+. Theorem 8.8 (Alternating Series Sum Estimation). If s = ( ) n b n is the sum of an alternating series that satisfies (i) 0 b n+ b n for all n and (ii) lim b n =0, n then s s n b n+. Proof. In the proof of the alternating series test, it was found that the subsequence {s 2k } approaches the limit value s from below, s 2k s. On the other hand, the subsequence {s 2k } approaches the limit value s from above. Indeed, s = b, s 3 = s b 2 + b 3 s because b 3 b 2, and, in general, s 2k+ = s 2k b 2k + b 2k+ s 2k ; that is, {s 2k+ } is monotonically decreasing. This shows that the sequence of partial sums s n oscillates around s so that the sum s always lies between any two consecutive partial sums s n and s n+ as depicted in Figure 8.8. Hence, s s n s n+ s n = b n+. Example Estimate the number of terms in a partial sum s n needed to approximate the sum of the alternating harmonic series correct to N decimal places. Solution: Here, b n =/n. Hence, the approximation s s n is correct to N decimal places if the absolute error does not exceed 5 0 N : s s n b n+ < 5 0 N or /(n+) < 5 0 N or n>0.2 (0 N ). Remark. If the monotonicity condition b n+ b n holds only if n N, the conclusion of Theorem 8.8 also holds only if n N. Indeed, in the notation from Remark 2, put t = ( ) n c n, where c n = b n+n. Let t n be a partial sum for the series ( ) n c n. Then s = s N +( ) N t and s n = s N +( ) N t n N+ for n N. Therefore, s s n = t t n N+ c n N+2 = b n+ for all n N Exercises. In () (5), determine whether the series converges or diverges (here p is real).

108 02 8. SEQUENCES AND SERIES ( ) n ( ) n n cos(nπ/2) () (2) (3) ln(n +3) n3 + n 4/5 n= n= n= ( ) n n ( ) n n ( ) n n 3 (4) (5) (6) (n 3/2 +) 2/3 (ln n) p 2 n n= n=2 n= ( ) n (ln n) p ( π (7) (8) ( ) n sin (9) n n) n=2 n= n= ( ) n (0) n +/n n () ( ) n ( n n ) n (2) n= n= n= ( ) n ( ) n (3) (4) (5) n + p n p n= n= n= n= ( ) n n 2 n 4 + ( ) n nn n! ( ) n (n 2 + n +) (2n +3) 2 In (6) and (7), find n for which the approximation by partial sums s s n is correct to N decimal places for the series. ( ) n n ( ) n n /3 (6) (7) 0 n n /3 +6 n= (8) Prove that the sum of the alternating harmonic series is ( ) n =ln2. n n= Hint: Show that a partial sum of the alternating harmonic series is s 2n = h 2n h n, where h n = a n +lnn and the sequence {a n } is defined in Exercise Then use the result of the latter exercise to prove that s n ln 2 as n Absolutely Convergent Series. 55. Ratio and Root Tests Definition 8.7 (Absolute Convergence). A series a n is called absolutely convergent if the series of absolute values a n is convergent. The absolute convergence is stronger than convergence, meaning that there are convergent series that do not converge absolutely. For example, the alternating harmonic series a n, a n =( ) n /n, is convergent, but not absolutely convergent because the series of absolute values a n =/n is nothing but the harmonic series /n, which is divergent (as a p series with p = ). On the other hand, the absolute convergence implies convergence.

109 55. RATIO AND ROOT TESTS 03 Theorem 8.9 (Convergence and Absolute Convergence). Every absolutely convergent series is convergent. Proof. For any sequence {a n }, the following inequality holds; 0 a n + a n 2 a n because a n is either a n or a n. It shows that the series b n, where b n = a n + a n, converges by the comparison test because 2 a n = 2 a n converges if a n converges absolutely. Hence, the series an = b n a n converges as the difference of two convergent series. Example Test the series [sin n 2 cos(2n)]/n 3/2 for absolute convergence. Solution: Making use of the inequality A + B A + B and the properties that sin x and cos x, one infers a n = sin n 2 cos(2n) n 3/2 sin n +2 cos(2n) n 3/2 3 n 3/2. The series of absolute values a n con-verges by comparison with the convergent p series 3 n 3/2 (here p =3/2 > ). So the series in question converges absolutely. Definition 8.8 (Conditional Convergence). A series a n is called conditionally convergent if it is convergent but not absolutely convergent. Thus, all convergent series are separated into two classes of conditionally convergent and absolutely convergent series. The key difference between properties of absolutely convergent and conditionally convergent series is studied in the next section Ratio Test. Theorem 8.20 (Ratio Test). Given a series a n, suppose the following limit exists: lim a n+ = c, n a n where c 0 or c =. If c<, then a n converges absolutely. If c>, then a n diverges. If c =, then the test gives no information.

110 04 8. SEQUENCES AND SERIES Proof. If c<, then the existence of the limit means that, for any ε>0, there is an integer N such that ε < a n+ a n+ c<ε = <c+ ε = q< for all n N. a n a n Note that since c is strictly less than, one can always take ε>0 small enough so that the number q = c + ε<. In particular, put n = N + k, where k 2. Applying the inequality a n+ <q a n consecutively k times, a N+k <q a N+k <q 2 a N+k 2 < <q k a N = a N q N q N+k. This shows that (8.3) a n <βq n, β = a N q N, for all n N. The series a n converges by comparison with the convergent geometric series βq n = β q n because q<. So a n converges absolutely. If c>, then there is an integer N such that a n+ / a n > or a n+ > a n 0 for all n N. Hence, the necessary condition for a series to converge, a n 0asn does not hold; that is, the series an diverges. If c =, it is sufficient, to give examples of a convergent and divergent series for which c =. Consider a p-series n p. One has a n+ n p c = lim = lim n a n n (n +) = lim p n (+/n) = p for any p. But a p-series converges if p> and diverges otherwise. Example Find all values of p and q for which the series n= np q n converges absolutely. Solution: Here a n = n p q n. One has a n+ (n +) p q n+ c = lim = lim n a n n n p q n = q lim n (+/n) p = q. So, for q < and any p, the series converges absolutely by the ratio test. If q = ±, the ratio test is inconclusive, and these cases have to be studied by different means. If q =, then a n = n p = /n p, which is a p-series that converges if p > or p<. Thus, the series converges absolutely for all p if q < and for p< ifq = ±. Note that, for p<0 and q =, the series conditionally converges (i.e., it is convergent but not absolutely convergent). In this case, it is a convergent alternating p-series ( ) n /n p (see Exercise ).

111 55. RATIO AND ROOT TESTS Root Test. Theorem 8.2 (Root Test). Given a series a n, suppose the following limit exists: n an = c, lim n where c 0 or c =. If c<, then a n converges absolutely. If c>, then a n diverges. If c =, then the test gives no information. Proof. If c<, then, as in the proof of the ratio test, the existence of the limit means that, for any c<q<, there is an integer N such that n an <q = a n <q n for all n N. This shows that the series a n converges by comparison with the convergent geometric series q n, 0 < q <. So a n converges absolutely. If c>, then there exists an integer N such that n a n > for all n N, and hence the condition a n 0asn does not hold. The series a n diverges. If c =, consider a p-series: n n p =( n n) p p = by Theorem 8.6. But a p-series converges if p> and diverges if p<. So the root test is inconclusive. Example Test the convergence of the series a n, where a n = [(2n 2 +5)/(3n 2 + 2)] n. Solution: Here a n = a n, and the absolute convergence is equivalent to the convergence. One has lim n n an = lim So the series converges. n 2n n 2 +2 = lim n 2+5/n 2 3+2/n 2 = 2 3 < Oscillatory Behavior of Sequences in the Root and Ratio Tests. Consider a sequence defined recursively by a = and a n+ = (sin n)a 2 n. An attempt to test the convergence of a n by the ratio test leads to the sequence c n = a n+ / a n = sin n which does not converge as it 2 oscillates between 0 and /2. Similarly, the sequence used in the root test may also exhibit oscillatory behavior and be nonconvergent, for example, a n =( sin 2 n)n so that c n = n a n = sin n. The ratio and 2 root tests, as stated in Theorems 8.20 and 8.2, assume the existence of the limit c n c. What can be said about the convergence of a series when this limit does not exist?

112 06 8. SEQUENCES AND SERIES To answer this question, recall that, in the proof of the ratio or root test, the existence of lim n c n = c<has been used only to establish the boundedness of the sequence c n q<for all n N, which is sufficient for the series a n to converge. But the boundedness property does not imply the convergence! Evidently, the boundedness condition holds in the above examples, c n = sin n < for all 2 2 n. Similarly, the existence of the limit value c> has only been used to show that n a n or a n for infinitely many n to conclude that the sequence {a n } cannot converge to 0 and hence a n diverges. If a n+ / a n for all n N, then again {a n } cannot converge to 0 (by the proof of the ratio test). Thus, the convergence of {c n } in the root or ratio test is not really necessary. Theorem 8.22 (Ratio and Root Tests Refined). Given a series n a n, put c n = a n+ / a n or c n = n a n. Then c n q< for all n N = a n converges. { n an for infinitely many n a n+ = a a n for all n N n diverges for some integer N Wider Scope of the Root Test. If the limit of a n+ / a n exists, n then so does the limit of a n and (8.4) lim n a n+ a n = lim n n a n. The converse is not true; that is, the existence of the limit of n a n does not generally imply the existence of the limit of a n+ / a n (the latter may or may not exist). Furthermore, if the sequence n a n does not converge, neither does a n+ / a n. A proof of these assertions is given in more advanced calculus courses. Thus, the ratio test has the same predicting power as the root test only if a n+ / a n converges. In general, the root test (as in Theorem 8.22) has wider scope, meaning that whenever the ratio test shows convergence, the root test does, too, and whenever the root test is inconclusive, the ratio test is, too. The subtlety to note here is that the converse of the latter statement is not generally true; that is, the inconclusiveness of the ratio test does not imply the inconclusiveness of the root test. The assertion can be illustrated with the following example. Consider a convergent series obtained from the sum of two geometric series in which the order

113 55. RATIO AND ROOT TESTS 07 of summation is changed: a n = = 3 2 n= k=0 2 k + 3 k=0 3 k = 3 2. where the sum of a geometric series has been used (Theorem 8.8). Now note that if n =2k is even, then a 2k =(/3) k, and a 2k =(/2) k if n =2k is odd. Take the subsequence of ratios for even n =2k, c 2k = a 2k+ /a 2k =(2/3) k /9. It converges to 0 as k. On the other hand, the subsequence of ratios for odd n =2k diverges: c 2k = (3/2) k as k. So the limit of c n does not exist; moreover, the ratio test (as in Theorem 8.22) fails miserably because c n is not even bounded. The series converges by the root test. Indeed, c 2k = 2k a 2k =/ 2 < and c 2k = 2k a 2k =/ 3 <. Although the sequence c n does not converge (it oscillates between / 3 and / 2), it is bounded, c n q =/ 2 < for all n, and hence the series converges by Theorem A similar example is given in Exercise Thus, the ratio test is sensitive to the order of summation, while this is not so for the root test When the Ratio Test Is Inconclusive. Theorem 8.23 (De Morgan s Ratio Test). Let a n be a series in which a n+ / a n as n. The series converges absolutely if ( lim n a ) n+ = b<. n a n The proof of this theorem is left to the reader as an exercise (see Exercise ). Consider the asymptotic behavior of the ratio c n = a n+ / a n as n. The theorem asserts that if c n behaves as c n +b/n for large n (i.e., neglecting terms of order /n p where p>), then the series a n converges if b<. For a p-series, the ratio test is inconclusive (see the proof of the ratio test). However, De Morgan s test resolves the inconclusiveness. Indeed, for large n, n p c n = (n +) = p (+/n) p p n. where the asymptotic behavior has been found from the linearization f(x) =(+x) p f(0) + f (0)x = px for small x =/n. So b = p and the series converges if b< orp>. This illustrates a basic technical trick to applying De Morgan s test. Suppose that there is a function f(x) such that a n+ / a n = f(/n).

114 08 8. SEQUENCES AND SERIES If f is differentiable at x = 0, that is, f(x) f(0) + f (0)x as x 0, then a n+ / a n = f(/n) f(0) + f (0)(/n) =+f (0)/n, and the series a n converges absolutely if f (0) <. Note that the property f(0) = follows from the inconclusiveness of the ratio test Exercises. In () (5), determine whether the series is absolutely convergent, conditionally convergent, or divergent (here p is real). () (4) (7) (0) (3) (5) n= n= ( ) n 3 n (2) p n (5) n! ( ) n (ln n) p n=2 n= n= n= n n 2 ( 2 3 )n (3) n= ( ) n 2 /n (6) n p n=2 ( ) n (ln n) p () n p (ln n) n (4) p n n! 5 8 (3n +2) (8) n= n= n= ( ) n n 4 n3 + (9) ( 2n 3 + n ) n (2) 3n (2n) n! n! p, p 0 n ( ) n n p+3 n= n= n! n! n n n= ( + n (6) For which integers p>0 is the series n= (n!)2 /(pn)! convergent? (7) (Estimating sums). Given a series a n with positive terms, put c n = a n+ /a n. Suppose that c n c<, that is, the series converges, an = s. Let s n be a partial sum. Prove that s s n a n+ c n+ if {c n } is a decreasing sequence, and s s n a n+ c if {c n } is an increasing sequence. Hint: Use the geometric series as in the proof of the ratio test to estimate the remainder s s n = a n+ + a n+2 +. (8) Prove De Morgan s ratio test. n= ) n 2

115 56. REARRANGEMENTS 09 Hints: Compare the series a n with the convergent p-series b n, where b n = A/n p and p = ( b)/2 > ifb<. Show that n(b n+ /b n ) b as n. Next, show that, by choosing the constant A one can always make a n <b n for all n. (9) Consider a geometric series with q =/2in which the order of terms is changed by swapping terms in each consecutive pair: a + a 2 + a 3 + = Test the convergence of this series using the root and ratio tests. 56. Rearrangements Here the difference between conditionally convergent and absolutely convergent series is further refined through the concept of rearrangement. Definition 8.9 (Rearrangement). Let {k n }, n =, 2,..., be an integer-valued positive sequence in which every positive integer appears only once (i.e., k n = k n if and only if n = n ). Given a series a n, put a n = a kn. The series a n is called a rearrangement of a n. For a finite sum, a rearrangement of its terms does not change the value of the sum. This is not generally so for convergent series. Consider an alternating p-series: ( ) n (8.5) a n = = n n= n= The series is convergent but not absolutely convergent (its sum is s = ln 2; see Exercise ). One of its rearrangements reads (8.6) a n = n= in which two positive terms are always followed by one negative. Let s n and s n be partial sums of (8.5) and (8.6), respectively. Put h n = + /2 + + /n (a partial sum of the harmonic series). Then s 2n = h 2n h n. Furthermore, s 3n = n 2 4 2n = h 4n 2 4 4n 2 h n = h 4n 2 h 2n 2 h n =(h 4n h 2n )+ 2 (h 2n h n )=s 4n + 2 s 2n.

116 0 8. SEQUENCES AND SERIES Taking the limit n in this equality, one finds s = s + s/2 =3s/2, where s and s are the sums of (8.5) and (8.6), respectively. Thus, a rearrangement of the series has changed its sum! This fact is not specific to the example considered but inherent in all conditionally convergent series. Terms of a conditionally convergent series occur with different signs (positive and negative). By regrouping positive and negative terms, it will be proved that the sum of a conditionally convergent series can be made any number or ±. The analysis begins by studying the properties of sums of positive and negative terms of a conditionally convergent series. Given a number x, put x ± =(x ± x )/2. The number x + = x if x>0 and x + = 0 otherwise. Similarly, x = x if x<0 and x =0 otherwise. Lemma 8.. Given a series a n, consider two series a + n and a n, where a ± n =(a n ± a n )/2 (the series of positive and negative terms). Then (i) If a n converges absolutely, then a + n and a n converge. (ii) If a n is conditionally convergent, then a + n and a n diverge. Proof. Let a n = s< and a n = t, where t< if a n converges absolutely and t = if it is conditionally convergent. Let s ± n be partial sums of a ± n, s n be partial sums of a n, and t n be partial sums of a n. Since s n s and t n t as n, one infers that a + n a n = a n a + n + a = s+ n s n = s n n = a n s + n + s = s+ s = s n = t n s + + s = t, where s ± are the limits of s ± n.if a n converges absolutely, then t< and hence s ± = (t ± s)/2; that is, both series a ± n converge. If an is conditionally convergent, then t =, and both sequences s ± n diverge. Theorem 8.24 (Riemann s Rearrangement Theorem). Let a n be a series that converges, but not absolutely. Then, for any c that is a real number or ±, there exists a rearrangement a n whose sequence of partial sums {s n} converges to c. Proof. Let p,p 2,... denote nonnegative terms of a n in the order in which they occur, and let q,q 2,... denote negative terms of a n in the order in which they occur. In the notation of Lemma 8., the series p n and a + n as well as q n and a n may only differ by zero terms (if some a n = 0). So the series p n and q n diverge. Consider the following rearrangement. Given a number c, take first

117 56. REARRANGEMENTS k terms p n, such that the number c lies between the partial sums s k = p + p p k and s k ; that is, k is defined by the condition s k p k <c<s k or c s k <p k. If c<0, then skip this first step. Next, take first m terms q n where m is the smallest integer such that s k +m = s k +q + +q m <c; that is, m is defined by the condition s k +m + q m >c>s k +m or c s k +m < q m. This can always be done because partial sums of p n can be larger than any number, while partial sums of q n can be smaller than any number owing to the divergence of these series. So s s n s k, n k, where c s k <p k, s k s n s k +m, k n k + m, where c s k +m < q k, Next, take k 2 next terms p n, where k 2 is the smallest integer such that s k +m +k 2 >c, and take m 2 next terms q n, where m 2 is the smallest integer for which s k +m +k +m 2 <c, and so on. At the nth step of the procedure, let n be the integer for which the last term in s n is p kn and let n 2 be the integer for which the last term in s n2 is q mn, that is, n 2 = n + m n. The partial sums of the constructed rearrangement oscillate about c, reaching local minima s n and local maxima s n : (8.7) s n s n s n2, n n n 2, c s n <p kn, c s n2 < q mn. By convergence of the series a n, a n 0asn 0. Hence, p n and q n also converge to 0 and so do the subsequences p kn 0 and q mn 0. Thus, all local maxima and minima of the sequence of partial sum {s n } converge to c by (8.7), which shows that s n c. Finally, if c = ±, one can take any divergent sequence c n (or ) and construct a rearrangement such that s k overshoots c and s k +m undershoots c, s k +m +k 2 overshoots c 2 and s k +m +k 2 +m 2 undershoots c 2, and so on. Obviously, this sequence of partial sum diverges. Absolutely convergent series have a drastically different property. Theorem 8.25 (Rearrangement and Absolute Convergence). If a series a n converges absolutely, then every rearrangement of a n converges, and they all converge to the same sum Proof. Let t n = a + a a n be a partial sum of the series of absolute values. The sequence {t n } converges to a number t by the hypothesis; that is, for any ε>0, there is an integer N such

118 2 8. SEQUENCES AND SERIES that t t n <εfor all n>n. Therefore, n a k = t n t N+ = t n t + t t N+ k=n+ t n t + t t N+ < 2ε. So, by taking N large enough, the sum of any number of terms a k, k>n, can be made smaller than any preassigned positive number. Let s n and s n be partial sums of a n and its rearrangement a n. One can take n>nlarge enough such that s n contains a, a 2,...,a N (i.e., the integers, 2,..., N are in the set of integers k,k 2,..., k n in the notations of Definition 8.9). Then the difference s n s n contains only terms a k with k>n(the terms a, a 2,..., a N are cancelled). Therefore, s n s n < 2ε for all n>n. If s n s and s n s, then s s < 2ε, which shows that s = s because ε>0 is arbitrary. Thus, an absolutely convergent series is much like a finite sum. The sum does not depend on the order in which the summation is carried out. In contrast, the sum of a conditionally convergent series depends on the summation order. This is the characteristic difference between these two classes of convergent series Strategy for Testing Series. It would not be wise to apply tests for convergence in a specific order to find one that finally works. Instead, a proper strategy, as with integration, is to classify the series according to its form. One should also keep in mind that a conclusion about the convergence of a series can be reached in different ways.. Special series. A series a n coincides with (or is a combination of or is equivalent to) special series such as a p-series, alternating p-series, geometric series, telescopic series, and so on. Their convergence properties are known. 2. Series similar to special ones. If a series a n has a form that is similar to one of the special series, then one of the comparison tests should be considered. For example, if a n is a rational or algebraic function (contains roots of polynomials), then the series should be compared with a p-series. 3. Necessary condition for convergence. It is is always easier to check the condition a n 0asn than it is to investigate the series a n for convergence. If the condition does not hold, the series diverges. 4. Alternating series. If a n =( ) n b n, b n 0, then the alternating series test is an obvious possibility.

119 56. REARRANGEMENTS 3 5. Ratio and root tests. Absolute convergence implies convergence. So, if the ratio or root test shows convergence, then the series in question converges absolutely. If these tests show divergence, then the series in question may still converge but not absolutely, and a further investigation is required. The root test is convenient for series of the form (b n ) n. The ratio test is convenient when a n involves the factorial n! or similar products of integers. The root test has a wider scope, but it is more difficult to use. The ratio test is often inconclusive if a n is a rational or algebraic function (c n = a n+ / a n ). In this case, the asymptotic behavior of c n is rather easy to find, c n +b/n as n, and then use De Morgan s test. 6. Series of nonnegative terms. If a n = f(n) 0 and the integral f(x) dx is easy to evaluate, then the integral test is effective. Also, it can be used in combination with the comparison test: a n f(n) and f(x) dx converges and so is a n, or f(n) a n and f(x) dx diverges and so is a n. Example Test the series (n +)/(n 2 + n +) for convergence. Solution: For large n, the leading terms of the top and bottom of the ratio are n and n 2, respectively. So a n /n asymptotically for large n. The series resembles the harmonic series, which diverges. It is natural, then, to try to prove the divergence of the series by comparing it with the harmonic series: n + n 2 + n + > n n 2 + n + n n 2 + n 2 + n = 2 2n. Thus, the series indeed diverges by comparison with the harmonic series. Example Test the series 3 n /(2 4 6 (2n)) for convergence. Solution: Each term a n involves a factorial-like product of integers, which suggests the use of the ratio test: a n+ a n = So, the series converges. 3 n+ 2 4 (2n) = (2n) (2n +2) 3 n 2n Example 8.3. Test the series sin(n 2 )e n3/2 Solution: One has a n = sin(n 2 ) e n3/2 e n3/2 e n. for convergence.

120 4 8. SEQUENCES AND SERIES The series e n converges by the integral test: e x dx =/e <. Hence, the series in question converges absolutely. Alternatively, the convergence of b n, where b n = e n3/2, can be established by the root test: n bn = e n/2 0 < asn. Example Put h n = +/2 + +/n (a partial sum of the harmonic series). Investigate the convergence of n p e qhn, where p q. Solution: The ratio test is inconclusive: a n+ = (n +)p e qhn+ (n +)p = e q(h n+ h n) = ( + n )p e q a n n p e qhn n p n+. To apply De Morgan s test, the asymptotic behavior of a n+ /a n has to be investigated. Put f(x) = ( + x) p exp( qx/( + x)) so that a n+ /a n = f(/n). Using the linearization near x =0,(+x) p +px, and exp( qx/(+x)) qx/(+x) qx, the asymptotic behavior is obtained: f(x) ( + px)( qx) +(p q)x = a n+ + p q a n n. Thus, the series converges if p q < orp < q. Of course, one could simply calculate f (0) = p q, but this is a bit more involved than the above procedure for finding the asymptotic behavior Exercises. In () (5), test the series for convergence or divergence (here p is real). () (4) (6) (8) () (4) n= n= n= n +2 n (2) ( ) n n 2n +3 (5) n= n! (3n ) n=2 n=2 (3) ( ) n ln n n p n p p 2n n= (2n +3) n n 3 5 (2n ) (7) ( ) (3n 2 +) n/ (2n) n= n= sin(/n) tan(πn +/n) (9) (0) (ln n) ln n n p ( n n + p ) n 2,p>0 (2) n= n= np n p n ( /n), p < (5) n n= n= n= n! ( n p 2 ) n (3) n= ( n p ),p 0 n! e np

121 57. POWER SERIES Power Series Definition 8.0 (Power Series). Given a sequence {c n }, the series c n x n = c 0 + c x + c 2 x 2 + c 3 x 3 + n=0 is called a power series in the variable x. The numbers c n are called the coefficients of the series. In general, the series will converge or diverge, depending on the choice of x. The power series always converges for x = 0 to the number c 0. Example For what values of x does the power series n=0 xn /n converge? Solution: By the root test, n xn n = x n n x as n. So the series converges for all <x< and diverges as x> or x<. The root test is inconclusive for x = ±. These values have to be investigated by different means. For x =, the power series becomes the harmonic series /n, which is divergent. For x =, the power series becomes the alternating harmonic series ( ) n /n, which is convergent. Thus, the power series converges if x [, ) and diverges otherwise. Given a number a, consider a power series in the variable y = x a: c n y n = c n (x a) n. n=0 n=0 It is also called a power series centered at a or a power series about a. Let S be the set of all values of x for which a power series in x converges and let S a be the set of all values of x for which the corresponding power series in (x a) converges. What is the relation between S and S a? Since the series are obtained from one another by merely shifting the value of the variable by a number a, x x a, the set S a is therefore obtained by adding the number a to every element of S: (8.8) x S a x a S = S a = {x x a S}. For example, the series n=0 (x 2)n /n converges if x 2 [, ) or x [, 3) and diverges otherwise by Example Thus, the problem of finding the set S a is equivalent to the problem of finding the set S.

122 6 8. SEQUENCES AND SERIES 57.. Power Series as a Function. Suppose that a power series in x converges on a set S. Then it defines a function on S: f(x) = c n x n, x S. n=0 The set S is called the domain of such a function. Functions defined by power series are most common in applications. Many of them have special notations (like elementary functions sin, cos, exp, etc.). Their properties are well studied. In what follows, it will be shown that familiar elementary functions such as sin x, cos x, and exp x, etc can also be represented as power series. Example Find the domain of the Bessel function of order 0 that is defined by the power series J 0 (x) = ( ) n 2 2n (n!) 2 x2n, n=0 where, by common convention, 0!=. Solution: Since a n = c n x 2n contains the factorial, the ratio test is more convenient: a n+ a n = x 2 c n+ c n = x 2 2 2n (n!) 2 2 2(n+) ((n + )!) 2 = x (n +) 2 0 as n. So the series converges for all x. Values of a function defined by a power series can be estimated by partial sums that are polynomials in the variable x: n f(x) f n (x) = c k x k = c 0 + c x + c 2 x c n x n. k=0 Thus, partial sums define a sequence of polynomials that converges to the function on S, f n (x) f(x) for all x S. The accuracy of the approximation is determined by the remainder R n (x) =f(x) f n (x). The accuracy assessment is discussed in Section Since the remainder R n (x) is a function on S, the error of the approximation is not generally uniform; that is, it depends on x Radius of Convergence. The set S on which a power series is convergent is an important characteristic and its properties have to be studied. Lemma 8.2 (Properties of a Power Series). (i). If a power series cn x n converges when x = b 0, then it converges whenever x < b.

123 57. POWER SERIES 7 (ii). If a power series c n x n diverges when x = d 0, then it diverges whenever x > d. Proof. If c n b n converges, then, by the necessary condition for convergence, c n b n 0asn. This means, in particular, that, for ε =, there exists an integer N such that c n b n <ε= for all n>n. Thus, for n>n, c n x n = c nb n x n = cn b n x n < x n. b n b b which shows that the series c n x n converges by comparison with the geometric series q n, where q = x/b and x/b < or x < b. Suppose that c n d n diverges. If x is any number such that x > d, then c n x n cannot converge because, by part (i) of the lemma, the convergence of c n x n implies the convergence of c n d n. Therefore, cn x n diverges. This lemma allows us to establish the following description of the set S. Theorem 8.26 (Convergence Properties of a Power Series). For a power series c n x n, there are only three possibilities: (i) The series converges only when x =0. (ii) The series converges for all x. (iii) There is a positive number R such that the series converges if x <Rand diverges if x >R. Proof. Suppose that neither case nor case 2 is true. Then there are numbers b 0 and d 0 such that the power series converges for x = b and diverges for x = d. By Lemma 8.2, the set of convergence S lies in the interval x d for all x S. This shows that d is an upper bound for the set S. By the completeness axiom, S has a least upper bound R = sup S. If x >R, then x S, and c n x n diverges. If x <R, then x is not an upper bound for S, and there exists a number b S such that b> x. Since b S, c n x n converges by Lemma 8.2. Theorem 8.26 shows that a power series converges in a single open interval ( R, R) and diverges outside this interval. The set S may or may not include the points x = ±R. This question requires a special investigation just like in Example So the number R is characteristic for convergence properties of a power series. Definition 8. (Radius of Convergence). The radius of convergence of a power series c n x n is a positive number R>0 such that

124 8 8. SEQUENCES AND SERIES the series converges in the open interval ( R, R) and diverges outside it. A power series is said to have a zero radius of convergence, R =0, if it converges only when x =0. A power series is said to have an infinite radius of convergence, R =, if it converges for all values of x. The ratio or root test can be used to determine the radius of convergence. Corollary 8.3 (Radius of Convergence of a Power Series). Given a power series c n x n, c n+ if lim = α = R = n c n α, if lim n c n = α = R = n α, where R =0if α = and R = if α =0. Proof. Put a n = c n x n in the ratio test (Theorem 8.20). Then a n+ / a n = x c n+ / c n x α. The series converges if x α <, which shows that R = /α. Similarly, using the root test (Theorem 8.2), n a n = x n c n x α <, which shows that R =/α. Remark. If the sequences in Corollary 8.3 do not converge, then Theorem 8.22 should be used, where a n = c n x n. Once the radius of convergence has been found and 0 <R<, the cases x = ±R have to be investigated by some other means (as the root or ratio test is inconclusive in this case) to determine the interval of convergence S of a power series. Example Find the radius of convergence and the interval of convergence of the power series c n x n, where c n =( q) n / n +and q>0. Solution: c n+ c n = qn+ n +2 n + q n n + = q n +2 = q +/n +2/n q = α. Therefore, R =/α =/q. Ifx = /q, then c n x n =( ) n / n += ( ) n b n. The sequence b n converges monotonically to 0 so that ( ) n b n converges by the alternating series test. If x = /q, then c n x n = / n +> / 2n, n. The p-series /n /2 diverges (p =/2 < ) so that / n + diverges by the comparison test. Thus, the interval of convergence is S =[ /q, /q).

125 57. POWER SERIES 9 Example Find the radius of convergence and the interval of convergence of the power series n 2 (x +) n /q n, where q>0. Solution: Put y = x+. If S is the interval of convergence of c n y n, where c n = n 2 /q n, then the interval of convergence in question is obtained by adding to all numbers in S according to the rule (8.8). By Corollary 8.3, n cn = q n n2 = q ( n n) 2 q = α. So R =/α = q. Ify = q, then c n y n = n 2, and the series n 2 diverges (a n = n 2 does not converge to 0). If y = q, then c n x n =( ) n n 2, and the series diverges because a n =( ) n n 2, does not converge to 0. The series converges only if y = x + <q, and hence the interval of convergence is x ( q,q ) (the interval ( q, q) shifted by ) Exercises. In () (2), find the radius of convergence and the interval of convergence of the power series. () (4) (6) n 3 x n (2) n=0 n=0 ( ) n 2 n ln n xn (5) n=2 n(x +) n (7) n=0 (8) (4x +) n n 2 n= (9) (0). n 2 x 2n 2 4 (2n) n= (2) 4 n (x +3)n n! n=0 2 n n 3 xn (3) n= n= ( n 2 ) n x 2n n= n p (x 2)n,p>0 ( ) n n2 + n 3 +3 x n (x )n 3 5 (2n ) n= (n!) k () (kn)! xn, k > 0 (integer) (3) Let p<qbe real numbers. Give examples of power series whose intervals of convergence are (p, q), [p, q], (p, q], and [p, q). (4) The Airy function is defined by the power series n=0 A(x) =+ x x x

126 20 8. SEQUENCES AND SERIES Find its domain. (5) A function f is defined by the power series f(x) =p + qx + px 2 + qx 3 + px 4 + qx 5 + ; that is, its coefficients c 2k = p and c 2k = q, where p and q are real. Find the domain of f and an explicit expression of f(x) (the sum of the series). (6) If f(x) = c n x n, where c n+4 = c n for all n 0, find the domain of f and a formula for f(x). (7) Power series c n x n and b n x n have the radii of convergence R and R 2, respectively. What is the radius of convergence of (c n + b n )x n? (8) Suppose that the radius of convergence of c n x n is R. What is the radius of convergence of c n x kn, where k>0 is an integer? 58. Representation of Functions as Power Series Consider a power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n. It is a geometric series with q = x 2, and therefore it converges for all q = x 2 < orx (, ). Using the formula for the sum of a geometric series, one infers that +x = 2 x2 + x 4 + = ( ) n x 2n for all <x<, n=0 This shows that the function /( + x 2 ) can be represented as a power series in the open interval (, ). Note that this representation is valid only in the interval of convergence of the power series despite the fact that the function /( + x 2 ) is defined on the entire real line. In general, one can construct a representation of a function by a power series in (x a) for some a. The interval of validity of this representation depends on the choice of a. Example Find a representation of /x as a power series in (x a), a>0, and determine the interval of its validity. Solution: Put y = x a. The function can be rewritten in a form that resembles the sum of a geometric series: x = a( + y/a) = ( y ) n ( ) n = a a a (x n+ a)n, x (0, 2a). n=0 n=0 n=0

127 58. REPRESENTATION OF FUNCTIONS AS POWER SERIES 2 The geometric series converges if q = y/a = y /a <, and hence this representation is valid only if a <y<aor a <x a<aor 0 <x<2a Differentiation and Integration of Power Series. The formula for the sum of a power series c n x n is often complicated and, in most cases, cannot even be found explicitly. How can functions defined by a power series be differentiated and integrated? If a function is a finite sum f(x) = u (x) + + u n (x), then the derivative is the sum of derivatives f = u + + u n and, similarly, the integral is the sum of integrals fdx= u dx + + u n dx. This is not generally true for infinite sums. As an example, consider a function defined by the series sin(nx) f(x) = u n (x) =. n 2 n= n= By comparison with a p-series, u n (x) = sin(nx) /n 2 /n 2, this series converges for all x because /n 2 converges. If the series is differentiated just like a finite sum, that is, term-by-term, u n(x) = cos(nx)/n, then the series u n(x) diverges for x =2πk for any integer k as the harmonic series /n. So f (2πk) does not exist. Thus, although the terms u n (x) are differentiable functions in the interval of convergence of the series u n, the series of derivatives u n may not converge and hence f = u n may not be differentiable everywhere in its domain. It appears that if u n (x) =c n (x a) n, that is, u n (x) isapower series, then the term-by-term differentiation or integration is justified. A proof of this assertion is beyond the scope of this course. Theorem 8.27 (Differentiation and Integration of Power Series). If the power series c n (x a) n has a nonzero radius of convergence R>0, then the function f defined by f(x) =c 0 + c (x a)+c 2 (x a) 2 + = c n (x a) n is differentiable (and therefore continuous) on the interval (a R, a+r) and f (x) =c +2c 2 (x a)+3c 3 (x a) 2 + = nc n (x a) n, (x a) 2 f(x) dx=c + c 0 (x a)+c + =C + 2 n=0 n= n=0 (x a) n+ c n. n +

128 22 8. SEQUENCES AND SERIES The radii of convergence of these power series are both R. Thus, for power series, the differentiation or integration and the summation can be carried out in any order: d cn (x a) n = d dx dx [c n(x a) n ], ( ) cn (x a) n dx = [c n (x a) n ] dx. Remark. Theorem 8.27 states that the radius of convergence of a power series does not change after differentiation or integration of the series. This does not mean that the interval of convergence does not change. It may happen that the original series converges at an endpoint, whereas the differentiated series diverges there. Example Find the intervals of convergence for f, f, and f if f(x) = n= xn /n 2. Solution: Here c n =/n 2 and hence n c n =/ n n 2 =(/ n n) 2 =α. So the radius of convergence is R =/α =. Forx = ±, the series is a p-series /n 2 that converges (p =2> ). Thus, f(x) is defined on the closed interval x [, ]. By Theorem 8.27, the derivatives f (x) = n= xn /n and f (x) = n=2 (n )xn 2 /n have the same radius of convergence R =. Forx =, the series f ( ) = ( ) n /n is the alternating harmonic series that converges, whereas the series f ( ) = ( ) n (n )/n diverges because the sequence of its terms does not converge to 0: ( ) n (n )/n = /n 0. For x =, the series f () = /n is the harmonic series and hence diverges. The series f () = (n )/n also diverges ((n )/n does not converge to 0). Thus, the intervals of convergence for f, f, and f are, respectively, [, ], [, ), and (, ). The term-by-term integration of a power series can be used to obtain a power series representation of antiderivatives. Example Find a power series representation for tan x. Solution: tan dx ( x = +x = ) ( x 2 ) n n+ x2n+ dx = C + ( ) 2 2n +. n=0 Since tan 0 = 0, the integration constant C satisfies the condition 0=C +0orC = 0. The geometric series with q = x 2 converges if q <. Hence, the radius of convergence of the series for tan x is R = (the power series representation is valid for x (, )). n=0

129 58. REPRESENTATION OF FUNCTIONS AS POWER SERIES 23 In particular, the number / 3 is less than the radius of convergence of the power series for tan x. So the number tan (/ 3) = π/6 can be written as the numerical series by substituting x =/ 3 into the power series for tan x. This leads to the following representation of the number π: π =2 3 n=0 ( ) n (2n + )3 n Power Series and Differential Equations. A power series representation is often used to solve differential equations. The relation between a function f(x), its argument x, and its derivatives f (x), f (x), and so on is called a differential equation. A function f(x) that satisfies a differential equation is generally difficult to find in a closed form. A power series representation turns out to be helpful. Since in this representation a function is defined by a sequence {c n }, f(x) = c n x n, and so are its derivatives f (k) (x), a differential equation imposes conditions on c n that are solved recursively. Example Find a power series representation of the solution of the equation f (x) =f(x) and determine its radius of convergence. Solution: Put f(x) = c n x n and hence f (x) = nc n x n. Then the equation f = f gives c +2c 2 x +3c 3 x 2 +4c 4 x 3 + = c 0 + c x + c 2 x 2 + c 3 x 3 +. By matching the coefficients at the monomial terms, x, x 2, x 3, and so on, one finds: c 0 = c c 2 = c 2, c 3 = c 2 3,..., c n = c n n. Using the latter relation recursively: c n = n c n = n(n ) c n 2 = n(n )(n 2) c n 3 = = c 0 n!. So f(x) =c 0 n=0 xn /n!, where c 0 is a constant (the equation is satisfied for any choice of c 0 ). By the ratio test, the series converges for all x (so R = ). Indeed, c n =/n! and c n+ /c n =/(n +) 0=α and hence R =/α =. For this simple differential equation, it is not difficult to find f(x) = c 0 e x by recalling the properties of the exponential function: (e x ) = e x. The condition f(0) = e 0 = determines the constant c 0 =. Thus,

130 24 8. SEQUENCES AND SERIES the exponential function has the following power series representation: (8.9) e x =+ x! + x2 2! + x3 3! + = n=0 x n n!. The series converges on the entire real line. In particular, the number e has the following series representation: e =+! + 2! + 3! + = n=0 n! Approximation of Definite Integrals. If an indefinite integral of f(x) is difficult to obtain, then the evaluation of the integral b f(x) dx a poses a problem. A power series representation offers a simple way to approximate the value of the integral. Suppose that f(x) = c n x n for R <x<r. By Theorem 8.27, for any R <a<b<r, b f(x) dx = b c n x n dx = b n+ c n a a n + a n+ c n n + n b k+ n c k k + a k+ c k k +. k=0 k=0 Errors of the approximation of the series sum by finite sums have been discussed earlier. Example 8.4. How many terms does one need in the power series approximation of the integral of f(x) =e x2 over the interval [0, ] to make the absolute error smaller than 0 5? Solution: Note first that the indefinite integral e x2 dx cannot be expressed in elementary functions! So a direct use of the fundamental theorem of calculus becomes problematic. However, e x2 dx can be represented as a power series that converges on the entire real line by replacing x in (8.9) by ( x 2 ). One has n 0 e x2 dx = k=0 ( ) k k! 0 x 2k dx = k=0 ( ) k k!(2k +) k=0 ( ) k k!(2k +). To determine n in the finite sum approximation of the series, recall the alternating series estimation theorem (Theorem 8.8), where b n = /(n!(2n + )): 0 e x2 dx n ( ) k b n+ = k!(2k +) k=0 (n + )!(2n +3) < 0 5.

131 58. REPRESENTATION OF FUNCTIONS AS POWER SERIES 25 A direct calculation shows that b and b So n = 7 is sufficient to approximate the integral with the required accuracy Exercises. In () (3), find a power series representation for the function and determine the interval of convergence. () f(x) = x 4 (2) f(x) = x 3x 2 +2 (3) f(x) = x + 2x 2 x In (4) (6), use differentiation to find a power series representation for the function and determine the interval of convergence. (4) f(x) = ( + x) 2 (5) f(x) = x 3 ( 4x 2 ) 2 (6) f(x) = ( + x 4 ) 3 In (7) (9), use integration to find a power series representation for the function and determine the radius of convergence. ( +x 2 ) (7) f(x) = ln(+x) (8) f(x) = ln (9) f(x) = tan (3x) x 2 In (0) (2), find a power series representation for the indefinite integral and determine the radius of convergence. ln( x) e x x (0) dx () dx (2) tan (x 2 ) dx x x 2 (3) Find a power series representation for sin x and cos x using the differential equation f +f = 0. Determine the interval of convergence. (4) Show that the Bessel function of order 0 defined in Example 8.34 satisfies the differential equation: x 2 J 0 (x)+xj 0(x)+x 2 J 0 (x) =0. In (5) (7), use differentiation or integration to find the sum of the series. (5) nx n (6) n= n(n )x n 2 (7) n=2 n=0 x n+ n + In (8) (20), how many terms does one need in a power series approximation to evaluate the integral with the absolute error not exceeding 0 6? (8) 0 dx +x 8 (9) 0 e x x dx (20) /2 0 ln( + x 4 ) dx

132 26 8. SEQUENCES AND SERIES (2) Find the radius of convergence of the hypergeometric series: + ab a(a +)b(b +) x+ x 2 a(a + )(a +2)b(b + )(b +2) + x 3 +,! c 2! c(c +) 3! c(c + )(c +2) where a, b, and c are reals. Use De Morgan s test to determine the interval of convergence. 59. Taylor Series 59.. Real Analytic Functions. Suppose a function f is represented by a power series (R >0): f(x) =c 0 + c (x a)+c 2 (x a) 2 + = c n (x a) n, x a <R. By Theorem 8.27, its derivatives f (k) (x) can obtained by the term-byterm differentiation of the series, and the resulting series has the same convergence radius R. Evidently, f(a) =c 0. What is the significance of the other coefficients c n? The derivative f is given by f (x) =c +2c 2 (x a)+3c 3 (x a) kc k (x a) k +, which shows that f (a) =c. The second derivative is f (x) =2c c 3 (x a)+ + k(k )c k (x a) k 2 +. Therefore, f (a) =2c 2. After k such steps, f (k) (x) =k(k ) 2 c k +(k +)k(k ) 2c k+ (x a)+, and hence f (k) (a) =k!c k or c k = f (k) (a)/k!. This proves the following theorem. Theorem 8.28 (Significance of Power Series Coefficients). If f has a power series representation n=0 f(x) = c n (x a) n, x a <R, n=0 for some a and R>0, then its coefficients are c n = f (n) (a). n! Definition 8.2 (Real Analytic Functions). A function f on an open interval I is said to be analytic if, for any a I, it has a power series representation f(x) = c n (x a) n that converges in some open interval (a δ, a + δ) I, where δ>0.

133 59. TAYLOR SERIES 27 The class of analytic functions plays a significant role in applications. Their properties are discussed next. Theorem 8.29 (Power Series Representation of Analytic Functions). A function f that is analytic on an open interval I has the power series representation (8.20) f(x) = n=0 f (n) (a) n! (x a) n for any a I that converges in an open subinterval of I that includes a. (8.2) This theorem follows from Definition 8.2 and Theorem In Example 8.37 it was found that x = n=0 ( ) n a n+ (x a)n, x (0, 2a). This shows that the function f(x) = /x is analytic for all x > 0 because a can be any positive number; that is, the function has a power series representation that converges in an open subinterval of (0, ) containing any a > 0. Similarly, the analyticity of f(x) = /x can be established for all x<0. It is important to emphasize that a power series for an analytic function does not necessarily converge on the entire domain of the function. But an analytic function can always be represented by a convergent power series in a neighborhood of every point of its domain. Equation (8.2) illustrates the point. Theorem 8.30 (Properties of Analytic Functions). (i) The sums and products of analytic functions are analytic. (ii) The reciprocal /f of an analytic function f is analytic if f is nowhere zero. (iii) The composition f(g(x)) of analytic functions f and g is analytic. (iv). Analytic functions are differentiable infinitely many times. A proof of properties (i) (iii) is given in more advanced calculus courses. Property (iv) follows from Theorem Its converse is not generally true; that is, there are functions that are differentiable infinitely many times at a point, but they cannot be represented by a power series that converges in an open interval that includes this point. As an example, consider the function f(x) =e /x2 if x 0 and f(0)=0.

134 28 8. SEQUENCES AND SERIES The function is continuous at x = 0 because lim x 0 e /x2 = lim u e u =0=f(0). It is differentiable at x = 0 because f f(x) f(0) (0) = lim x 0 x e /x2 = lim x 0 x =0. The first equality is the definition of f (0). The last limit is established by investigating the left and right limits x 0 ± with the help of the substitution x =/u ± as x 0 ± ; the left and right limits coincide because ue u2 0asu ± (the exponential function decreases faster than any power function). In a similar fashion, it can be proved that f (n) (0) = 0 for all n (see Exercise ). Thus, f(x) has no power series representation c n x n in a neighborhood of x =0 because, if it did, then, by Theorem 8.29 the function should have been identically 0 in some interval ( δ, δ), δ>0, (as f (n) (0) = 0 for all n), which is not true (f(x) 0 for all x 0). Hence, the function is not analytic at x = Taylor and Maclaurin Series. Definition 8.3 (Taylor and Maclaurin Series). The series in (8.20) is called the Taylor series of a function f at a (or about a, or centered at a). The special case of the Taylor series when a =0is called the Maclaurin series of a function f. The Taylor series of the exponential function e x about x =0is given by (8.9). The series converges for all x; that is, its radius of convergence is R =. Trigonometric functions. Consider the Maclaurin series of f(x) = sin x. One has f (x) = (sin x) = cos x and f (x) = (cos x) = sin x. Hence, f (2n) (x) =( ) n sin x, f (2n+) (x) =( ) n cos x, and f (2n) (0)=0,f (2n+) (0)=( ) n.so sin x = n=0 ( ) n x 2n+ (2n + )! = x x3 3! + x5 +, R =. 5! By the ratio test, c n+ / c n =(2n + )!/(2n +3)!=/[(2n + 2)(2n + 3)] 0=α, and the radius of convergence is R =/α =. The series converges on the entire real line. The Maclaurin series for f(x) = cos x is obtained by differentiating cos x = (sin x) = ( ) n x 2n n=0 (2n)! = x2 2! + x4 +, R =. 4!

135 59. TAYLOR SERIES 29 By Theorem 8.27 it also converges on the entire real line. Binomial series. Let f(x) =(+x) p, where p is any real number. Its derivatives are f (x) =p( + x) p, f (x) =p(p )( + x) p 2, and, in general, f (n) (x) =p(p ) (p n + )( + x) p n. The Maclaurin series for ( + x) p is called the binomial series. The traditional notation for its coefficients is ( p = =. n) c n = f (n) (0) n! p(p ) (p n +) n! These numbers are called the binomial coefficients. The binomial series and its radius of convergence are ( p ( + x) p = x n) n =+ p p(p ) x + x 2 +, R =.! 2! n=0 The coefficients satisfy the recurrence relation c n+ = c n (p n)/(n+). Therefore, by the ratio test, c n+ / c n = p n /(n+) = p/n /(+ /n) =α as n. Hence, R =/α = Taylor Series of Analytic Functions. Every analytic function in a neighborhood of any point is represented by the Taylor series about that point. If the Taylor series converges on the entire real line, then the function is analytic everywhere. In particular, the exponential e x and trigonometric functions sin x and cos x are analytic everywhere. Moreover, the properties of analytic functions stated in Theorem 8.30 allows us to add, multiply, and make a composition of the Taylor series (on the common intervals of their convergence) just like ordinary sums to obtain the Taylor series representation of the sums, products, and compositions of analytic functions. These are extremely useful properties in applications. Example Find first four terms of the Taylor series for the function f(x) = exp(tan x) about x =0. Solution: Calculation of the derivatives of such a function is rather tedious. Instead, note that e x and tan x are both analytic in a neighborhood of x = 0. So the composition of their Taylor series (see (8.9) and Example 8.39) gives the sought-after Taylor series. Only monomials, x, x 2, and x 3 have to be retained when calculating the composition. This implies that it is sufficient to retain two leading terms in

136 30 8. SEQUENCES AND SERIES the Taylor series tan x = x x 3 /3+ and four leading terms in the Taylor series (8.9) of the exponential function: e tan x = + tan x + ( ) 2 tan ( 3 x + tan (x)) =+ (x x3 =+ (x x3 3 (x ) x ) ) + 2 x2 + 6 x3 + =+x + 2 x2 6 x3 +. (x x3 3 + ) Approximations by Taylor Polynomials. An analytic function f can be approximated by a finite sum of the Taylor series: n f (k) (a) f(x) (x a) k = T n (x). k! k=0 The polynomial T n (x) is called the Taylor polynomial about a. The convergence of the Taylor series guarantees that the remainder converges to 0: R n (x) =f(x) T n (x) 0 as n for all x a <R, where R is the radius of convergence of the Taylor series. The accuracy of the Taylor polynomial approximation for a function is assessed in Taylor s theorem discussed in Calculus I. Here it is restated in a slightly different form. Theorem 8.3 (Taylor s Theorem). Suppose a function f is analytic near a and let T n (x) be its Taylor polynomials about a. Then, for every n and any x a <R, where R is the radius of convergence of the Taylor series for f about a, there exists a point ξ between a and x such that R n (x) =f(x) T n (x) = f (n+) (ξ) (n + )! (x a)n+. Proof. Given a number x, x a <R, let M be a number defined by f(x) =T n (x)+m(x a) n+. Consider the function g(t) =f(t) T n (t) M(t a) n+, where t a <R.

137 59. TAYLOR SERIES 3 Since the (n + )th derivative of a polynomial of degree n vanishes and hence g (n+) (t) =f (n+) (t) n!m, the proof will be complete if one can show that g (n+) (ξ) = 0 for some ξ between x and a (the latter would imply that M = f (n+) (ξ)/n!). By the definition of Taylor polynomials, f (k) (a) =T n (k) (a) for k =0,,..., n, and hence g(a) =g (a) = = g (n) (a) = 0. The function g(t) is differentiable and g(x) =g(a) =0 by the choice of M; therefore, by Rolle s theorem, there is a number t between x and a such that g (t ) = 0. Similarly, the function g (t) is differentiable and g (t )=g (a) = 0; hence, there is a number t 2 between t and a such that g (t 2 ) = 0. After n + steps of this procedure, one arrives at the conclusion that g (n+) (t n+ ) = 0 for some number t n+ = ξ between t n and a, that is, between x and a. Corollary 8.4 (Taylor s Inequality). If f (n+)(x) M n for x a d < R, then the remainder of the Taylor series satisfies the inequality R n (x) M n (n + )! x a n+ for x a d<r. Since ξ in Taylor s theorem lies between x and a, one has f (n+) (ξ) M n for x a d, and the conclusion of the corollary follows. All derivatives of an analytic function are continuous and, hence, attain their maximal and minimal values on any closed interval x a d. So M n = max f (n+) (x) on x a d. Example Find an upper bound on the error of the Taylor polynomial approximation about x =0for the function f(x) = sin x. Solution: The Maclaurin series for sin x contains only odd powers of x and so are the Taylor polynomials. Since f (2n+2) (x) =( ) n+ sin x, one has f (2n+2) (x) = sin x =M 2n+ uniformly for all x and all n and hence sin x T 2n+ (x) x 2n+2 (2n + 2)!. This example shows that, although the error converges to 0 for all x, for a fixed n it grows with increasing x. This implies that Taylor polynomials of higher degrees are needed to achieve the same accuracy for large x as for smaller x (see Fig. 8.9). To avoid using high-degree Taylor polynomials to approximate the function at large x, one can use Taylor polynomials about some a close to the range of x in which the approximation is needed.

138 32 8. SEQUENCES AND SERIES Figure 8.9. An illustration of an approximation of f(x) = sin x (the dashed red curve) by its Taylor polynomials at x = 0 (the solid blue curve). As n increases, T n (x) approaches f(x) = sin x. The approximation becomes better in a larger interval for a larger n in accordance with the analysis of Example Exercises. In () (5), find the Maclaurin series for the function and the radius of convergence. () ln(+x) (2) tan x (3) sinh x (4) cosh x (5) x 6 +2x 5 x 3 +x 3 In (6) (9), find the Taylor series for the function about a and the radius of convergence. (6) cos x, a= π (7) / x, a=4 (8) sin x, a= π/2 (9) ( + x) 2/3,a=7 In (0) (3), use Maclaurin series for basic functions to find the Maclaurin series for the function. (0) x cos(x 2 /2) () x 3 +x 4 (2) x sin x x 3 (3) x 2 tan (x 2 )

139 59. TAYLOR SERIES 33 In (4) (7), use the products and composition of the Maclaurin series for basic functions to find the first three non-vanishing terms of the Maclaurin series for the function. (4) sin(π(cos x)) (5) e sin x (6) tan x ln( + x) (7) ln(cos x) (8) Find the first five nonvanishing terms of the Maclaurin series for f(x) =e x / cos x. Hint: Put f(x) =c 0 + c x + + c 4 x 4 +, then use the product of the Maclaurin series to find the coefficients from e x = f(x) cos x. In (9) (2), find the degree of a Taylor polynomial to approximate the integrand so that the error of approximating the integral does not exceed 0 4. /2 (9) tan (x 2 e x /2 ) dx (20) dx (2) ( + x 4 ) /4 dx 0 0 x 0 In (2) (23), find the sum of the series. (2) n=0 ( ) n π 2n 6 2n (2n)! (22) ( ) n 3 n n=0 2 n n! (23) ln 2+ (ln 2)2 (ln 2)3 + 2! 3! (24) (i) For the function f(x) =e /x2 if x 0 and f(0) = 0, show that f (n) (0) = 0 for all n and hence f cannot be represented as a power series near 0. (ii) Let f(x) =e /x if x>0 and f(x) =0ifx 0. Is this function analytic everywhere?

140

141 CHAPTER 9 Further Applications of Integration 60. Arc Length 60.. The Length of a Curve. We have seen various applications of integration to the computation of the area of a domain and to the computation of the volume of a solid. It is perhaps more surprising that we can also use integration to compute the length of a curve between two given points. This may sound counterintuitive at first, since in the applications we have seen so far, integration was used to compute some parameter of an object that existed in a higher dimension than the function that was being integrated. Let f be a function so that, on the interval [a, b], the derivative f of f exists and is a continuous function. We would like to know the length of the curve of f, starting at the point A =(a, f(a)) and ending at the point B =(b, f(b)). Intuitively, we can imagine that we lay a rope over the graph of f between the two endpoints, mark A and B on the rope, then straighten that rope out, and measure the distance between them. A more formal definition, which is useful in the actual computation of the length of the curve, is the following. Cut the interval [a, b] into n equal parts, using points a = x 0 <x < <x n = b. Let P i = (x i,f(x i )) = f(x i,y i ). Let P i P i denote the length of the straight line segment from P i to P i. Then the sum n (9.) K n = P i P i i= is a little bit smaller than the length of the curve since the points P i and P i are on the curve and the straight line is the shortest path between them. If we keep refining the subdivision of the interval [a, b] by having n go to infinity, then it can be proved that lim n K n exists. We define that limit to be the length of the curve of f from A to B. See Figure 9. for an illustration. Note that in the case when the graph of f is a straight line segment between A and B, this definition is just the length of that segment, so our definition extends our previous notion of length. 35

142 36 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.. Arc length as a limit. Let us now return to (9.) in order to compute lim n K n. Let (b a)/n =(x i x i )/n = x. Note that then P i P i = (x i x i ) 2 +(y i y i ) 2 = ( x) 2 +( x) (y 2 i y i ) 2 (x i x i ) 2 = x + (y i y i ) 2 (x i x i ). 2 Now observe that since f is continuous, the intermediate value theorem implies that there is a real number x i [x i,x i ] such that = f (x i ). Hence, the previous chain of equalities yields y i y i x i x i Summing over all i, weget P i P i = x +f (x i )2. K n = n x +f (x i )2. i= As n goes to infinity, the left-hand side, by definition, converges to the length of the curve of f between A and B, while the right-hand side, being a Riemann sum, converges to b a +f (x) 2 dx. Hence, we have proved the following theorem. Theorem 9.. If f is a continuous function on the interval [a, b], then the length of the graph of f(x) from the point (a, f(a)) to the point

143 60. ARC LENGTH 37 Figure 9.2. The curve of f(x) = 2 3 x3/2. (b, f(b)) is equal to L = b a +f (x) 2 dx. Example 9.. Find the length of the curve of f(x) = 2 3 x3/2 from (0, 0) to (, 2/3). See Figure 9.2 for an illustration. Solution: We have f (x) = x,sof is a continuous function on [0, ], and therefore Theorem 9. applies. Using that theorem, we obtain L = +( x) 2 dx = = 0 0 +xdx [ 2 ( + x)3/2 3 = Note that the result is remarkably close to the length of the straight line that connects the two points in question, which is 3/3. We can use our new technique to verify a classic formula. Example 9.2. Use Theorem 9. to compute the circumference of a circle of radius. Solution: Let us place the center of the unit circle at the origin. Then the boundary of the circle is the set of points satisfying x 2 + y 2 =. ] 0

144 38 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.3. One quarter of the unit circle. We want to use Theorem 9., so we need a part of the circle where that satisfies the vertical line test (so y is a function of f) and where the tangent line to the circle is never vertical (so that f (x) exists). For instance, ( we can choose the quarter of the circle that starts in the point ) ( 2 2 and ends in the point 2 2 ). See Figure 9.3 for an illustration. On that part of the curve, f (x) = x x 2 is continuous, so Theorem 9. implies L = = = 2/2 2/2 2/2 2/2 2/2 2/2 = [sin x] = π/2. +f (x) 2 dx + x2 x 2 dx x 2 dx 2/2 2/2 This implies that the circumference of the full circle is four times this much, that is, 2π.

145 6. SURFACE AREA Remarks. Recall that in the first paragraph of this section, we discussed why it may seem counterintuitive that integration plays a role in the computation of arc lengths. Now we can see that the purported contradiction explained there is resolved by the fact that the integrand in Theorem 9. contains f, not f. Compared to other formulas we learned in our earlier studies of integration, it is relatively rare that the formula given by Theorem 9. can be explicitly computed, since b a +f (x) 2 dx is often difficult to handle. Therefore, we must often resort to approximate integration while computing arc lengths Exercises. () Find the length of the curve f(x) =x 2 /2 between the points given by x = 0 and x =. (2) Find the length of the curve f(x) =x between the points 4x given by x = 2 and x =4. (3) Find the length of the curve f(x) = ln(cos x) between the points given by x = 0 and x = π/4. (4) Prove that Theorem 9. provides the correct value for the arc length of f when f is a linear function. (5) Use a method of approximate integration to estimate the length of the curve of f(x) =e x as from (0, ) to (,e). (6) Use a method of approximate integration to estimate the length of f(x) = sin x from (0, 0) to (π, 0). 6. Surface Area 6.. The Definition of Surface Area. In the last section, we defined the length of a curve, and deduced a formula for the computation of that length. Let us now take a curve, say of a function f(x) =y, where x [a, b] and f is continuous on [a, b]. Let us rotate this curve around the horizontal axis, as shown in Figure 9.4. What is the area of the obtained surface of revolution? The definition of the area in question, and its computation, will be quite similar to what we have discussed in the previous section for the arc length. Cut the interval [a, b] inton equal parts, using points a = x 0 <x < <x n = b. Let P i =(x i,f(x i )) = f(x i,y i ) and let l i = P i P i denote the length of the straight line segment from P i to P i. As we rotate the curve of f around the horizontal axis, the rotation of the segment P i P i

146 40 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.4. A surface obtained by rotating a curve. Figure 9.5. Approximating a surface of revolution. results in the lateral surface S n,i of a truncated cone with slant height l i and radii y i and y i. See Figure 9.5 for an illustration. It is then not difficult to prove that the area of S n,i is equal to (9.2) A(S n,i )=π(y i + y i )l i. As n goes to infinity, the sum of the areas of the surfaces S n,i approximates what we intuitively think of as the area of the surface obtained by rotating the curve.

147 6. SURFACE AREA 4 In fact, it can be proved that the limit (9.3) S(A) = lim n n i= S n,i = lim n n π(y i + y i )l i i= exists. We define this limit to be the area of the surface of revolution obtained when the curve of f is rotated around the horizontal axis, with x [a, b]. In order to compute this surface area, recall from the last section that there exists a real number x i [x i,x i ] such that l i = P i P i = x +f (x i )2. Also note that, since f is continuous, small changes in x lead to small changes in f(x) =y, soifn is large enough, then f(x i ) f(x i ) = y i and f(x i ) f(x i ) = y i. Therefore, (9.3) implies (9.4) S(A) = lim n n 2πf(x i ) x +f (x i )2, i= where x =(b a)/n. Now notice that the last expression obtained for S(A) is the limit of a Riemann sum, that is, an integral (of the function 2πf(x) +f (x) 2 ). This means that we proved the following theorem. Theorem 9.2. Let f be a function that f is continuous on the interval [a, b]. LetS be the surface obtained by rotating the curve y = f(x), where x [a, b], around the horizontal axis. Then the area of S is A(S) = b a 2πf(x) +f (x) 2 dx. Example 9.3. Compute the surface area of a sphere of radius r. Solution: Such a sphere can be obtained by rotating the semicircle given by the equation f(x) = r 2 x 2 around the horizontal axis. See Figure 9.6 for an illustration.

148 42 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.6. A sphere as a surface of revolution. Theorem 9.2 then yields A(S) = = = r r r r r r 2π r 2 x 2 r 2 x dx 2 2π r 2 r 2 x 2 r 2 x dx 2 2rπ dx =[2rπx] r r =4r 2 π. + x Variations. If we rotate our curve around the vertical axis instead of the horizontal axis, then most of the previous argument remains valid. The only difference is that when the point P i =(x i,y i ) is rotated, it is rotated in a circle of radius x i, not y i. This leads to the following theorem. Theorem 9.3. Let f be a function such that f is continuous on the interval [a, b]. Let S be the surface obtained by rotating the curve y = f(x), where x [a, b], around the vertical axis. Then the area of S is A(S) = b a 2πx +f (x) 2 dx. Note that the f(x) term in the integrand of Theorem 9.2 is replaced by x.

149 6. SURFACE AREA 43 Example 9.4. Rotate the curve given by y = f(x) =x 2 /2, with x [0, ], around the vertical axis. Find the area of the obtained surface. Figure 9.7. The curve of y = x 2 /2 and the surface obtained by its rotation. Solution: Theorem 9.3 implies A(S) = 0 =2π 2πx +x 2 dx [ ] 3 (x2 +) 3/2 =2π Note that another way of writing the result of Theorem 9.2 is b ( ) 2 dy A(S) = 2πf(x) + dx. a dx By interchanging the roles of x and y, this implies that, for curves given by an equation g(y) = x, the following holds. Theorem 9.4. Let g be a function such that g is continuous on the interval [a, b]. Let S be the surface obtained by rotating the curve x = g(y), where y [a, b], around the vertical axis. 0

150 44 9. FURTHER APPLICATIONS OF INTEGRATION Then the area of S is given by b A(S) = 2πg(y) + a ( ) 2 dx dy. dy While the surface area of a cone can be computed by elementary methods, it is elucidating to compute it with our new method and see that the result is what we expect it to be. Example 9.5. Find the surface area of a right cone of base radius R and height h. Solution: The base circle of the cone has area R 2 π. In order to compute the lateral surface, note that the lateral surface can be obtained by rotating the line segment given by the equation x = Ry + R = g(y), h where y [0,h], around the vertical axis. Therefore, Theorem 9.4 applies, and for the lateral surface area, it yields h ( A(S) = 2π Ry ) 0 h + R + R2 h dy 2 ] R2 + h =2πR 2 h [ y2 h 2 2h + y = πr R 2 + h 2 = πrs, where s = R 2 + h 2 is the slant height of the cone. So the total surface area of the cone is the sum of the area of its base plus the area of its lateral surface, that is, R 2 π+πrs = πr(r+s). Theorem 9.4 also has a version that applies to curves given as functions of y that are rotated around the horizontal axis. Theorem 9.5. Let g be a function such that g is continuous on the interval [a, b]. Let S be the surface obtained by rotating the curve x = g(y), where y [a, b], around the horizontal axis. Then the area of S is given by b ( ) 2 dx A(S) = 2πy + dy. dy a Note that for the computation of some surface areas, we will have a choice of two theorems discussed in this section. The reader is encouraged to describe the curves whose rotations lead to such surfaces. 0

151 62. APPLICATIONS TO PHYSICS AND ENGINEERING Exercises. () Compute the surface area obtained by rotating the curve y = e x, for x [0, ], around the horizontal axis. (2) Rotate the curve of y = f(x) = x, where x [0, ], around the vertical axis. Find the surface area. (3) Rotate the curve of y = f(x) = tan x, where x [0,π/4], around the vertical axis. Find the surface area. (4) Rotate the curve of y = f(x) =x 3, where x [0, ], around the horizontal axis. Find the surface area. (5) Rotate the curve of x = g(y) = 2 3 (y +)3/2, where y [2, 3], around the horizontal axis. Find the surface area. (6) Solve Example 9.5 using Theorem Center of Mass. 62. Applications to Physics and Engineering One-Dimensional Systems. Let us assume that we have two objects of mass m and m 2 placed on the line of real numbers, at points x and x 2, respectively. We want to find the point x g such that if we place a fulcrum under the interval [x,x 2 ]atx g, the objects at the endpoints of the interval will balance. See Figure 9.8 for an illustration. We assume that the interval [x,x 2 ], or the stick representing it, has negligible mass. If m = m 2, then we clearly have x g =(x + x 2 )/2. Otherwise, we make use of the well-known fact of physics that the interval will balance if the moments on the two sides of the fulcrum are equal, that is, when (9.5) m (x g x )=m 2 (x 2 x g ) holds. Solving (9.5) for x g,weget (9.6) x g = m x + m 2 x 2 m + m 2. Figure 9.8. Center of mass.

152 46 9. FURTHER APPLICATIONS OF INTEGRATION The point x g of the real line is called the center of mass or center of gravity of the system described above, that is, the system of an object of mass m at x and an object of mass m 2 at x 2. The moment of an object with respect to a point P is the mass of the object times the distance of the object from P. In particular, in the above system, the two objects had moments m x and m 2 x 2 with respect to the origin. So the total system had moment m x + m 2 x 2. Note that if we replace the two objects by a simple object of mass m + m 2 placed at x g, then the moment of the system about the origin does not change. This is an important property that only the center of mass has, and therefore we repeat it. If we concentrate the total mass of the system at the center of mass, the moment of the system with respect to the origin will not change. If we consider a system of k distinct objects of mass m,m 2,...,m k placed at points x,x 2,...,x k along the horizontal axis, then we can use an analogous argument to show that the center of mass of the system is at (9.7) x g = k i= m ix i k i= m. i Two-Dimensional Systems Discrete Two-Dimensional Systems. Let us now consider the more general case when the k objects of mass m,m 2,...,m k are placed in points (x,y ), (x 2,y 2 ),...,(x k,y k )oftheplane. We would like to find the center of mass (x g,y g ) of this system. In other words, we assume that a plate of negligible mass is placed under our system, and we want to find the point (x g,y g ) with the property that if we place a fulcrum under the plate at that point, the plate will balance. Using methods similar to the one-dimensional case, it can be proved that the plate will balance if the fulcrum is placed at (x g,y g ) with (9.8) x g = k i= m ix i k i= m i and y g = k i= m iy i k i= m. i This corresponds to the intuitively appropriate concept that the plate will balance if it balances both horizontally and vertically. The sum M x = k i= m iy i is called the moment of the system with respect to the x axis. This name is due to the fact if we tried to balance the system on the x axis, the larger the number M x, the more would the weights of the system rotate the plate. Similarly, the sum

153 62. APPLICATIONS TO PHYSICS AND ENGINEERING 47 M y = k i= m ix i is called the moment of the system with respect to the y axis Symmetry Lines. Now let us consider the continuous version of the problem. Let P be a plate and let us try to find the center of mass of P. (We no longer assume that the mass of the plate is negligible; in fact, that mass is the object of our study now.) Let us assume, for the rest of this chapter, that the mass of P is uniformly distributed over P. Let us also assume that the density of the material of which P is made is. That is, the mass of a unit square within P is. Sometimes we can find the center of mass of P without computation. A symmetry line of P is a straight line t such that the image of P when reflected through t is P itself. That implies that the two parts into which t cuts P are congruent, and the plate balances on the line t. Consequently, the center of mass C of P must be on t, since if we concentrate the entire mass of P in C, it still has to balance on the line P. The argument of the previous paragraph shows that the center of mass of P must be on every symmetry line of P. So if P has more than one symmetry line, then these symmetry lines must all intersect in one point, namely, in the center of mass of P. In this case, we obtain the center of mass of P as the intersection of any two symmetry lines of P. For example, we can find the center of mass of a circle, ellipse, rectangle, or rhombus in this way A Formula for Continuous Two-Dimensional Systems. Let us keep the conditions from the previous section and let us impose the new condition that P is a domain under a curve ; that is, the borders of P are the vertical lines x = a and x = b, the horizontal axis, and the graph of the continuous function f(x) = y. We would like to use formula (9.8) to find the approximate location of the center of mass of P. Let us cut the interval [a, b] inton equal parts, using the intermediate points a = x 0 <x <x 2 < <x n = b, and let x =(b a)/n. The vertical lines x = x i cut P into n vertical stripes. Let S i be the ith such stripe. The area, and hence the mass, of S i is close to x f(x i ), where x i is the midpoint (x i + x i )/2 of the interval [x i,x i ]. So we are approximating S i by a rectangle R i. Let us concentrate the entire mass of R i in the center of mass of R i, that is, at (x i,f(x i )/2). Now we can compute the moment of the obtained system of n objects with respect to the x axis. Note that the mass of R i is equal to

154 48 9. FURTHER APPLICATIONS OF INTEGRATION the area of R i, that is, xf(x i ). So we have n n M x (n) = m i y i = x f(x i ) f(x i )/2. i= i= The right-hand side is a Riemann sum, so, as n goes to infinity, it will converge to the corresponding integral, while the left-hand side will converge to the moment M x of the original plate with respect to the x axis. This yields b (9.9) M x = a 2 f(x)2 dx. A similar argument using horizontal stripes instead of vertical ones shows that (9.0) M y = b a xf(x) dx. Finally, now that the moments of P are known, it is straightforward to compute the coordinates of the center of mass of P. Indeed, the center of mass is the unique point (x g,y g ) with the property that if the entire mass A = b f(x) dx of P is placed in that point, then the a moments of this one-object system are identical to the moments of P. In other words, M x = y g A and M y = x g A. Therefore, x g = M y /A and y g = M x /A, which means that formulas (9.9) and (9.0) imply the following theorem. Theorem 9.6. Let f be a function such that f(x) 0 if x [a, b]. Let D be a domain whose borders are the vertical lines x = a and x = b, the horizontal axis, and the curve of the function f(x) =y. LetA(D) denote the area of D. Let x g and y g be the coordinates of the center of mass of D. Then we have x g = b xf(x) dx a A(D) and y g = b a 2 [f(x)]2 dx A(D) Example 9.6. Find the coordinates of the center of mass of the quarter of the unit circle that is in the northeastern quadrant. Note that if we asked the same question for the entire unit circle, the answer would obviously be that x g = y g = 0, since the center of gravity of any domain must be on all symmetry lines. If we asked the same question for the half of the unit circle that is in the northern.

155 62. APPLICATIONS TO PHYSICS AND ENGINEERING 49 half-plane, then x g = 0 would clearly hold, since the vertical axis is a symmetry line of that semicircle. Solution: (of Example 9.6): Note that the domain D in question has a symmetry line, namely, the line determined by the equation x = y. So the center of gravity of D is on that line, that is, x g = y g. Therefore, it suffices to compute one of x g and y g. We have f(x) = x 2 = y, so y g is somewhat easier to compute. Theorem 9.6 yields y g = 0 ( 2 x2 ) dx π/4 = 2 ] [x π x3 3 0 = 2 π 2 3 = 4 3π. So the center of gravity of the quarter of the unit circle in the northeastern quadrant is at ( 4, 4 ). See Figure 9.9 for an illustration. 3π 3π Note that 4/(3π) This makes perfect sense since this shows that the center of gravity of D is closer to the horizontal axis (the bottom of the quarter circle) than to the y = line (the top of Figure 9.9. The center of mass of a quarter of the unit circle.

156 50 9. FURTHER APPLICATIONS OF INTEGRATION the quarter circle). That is reasonable, since the bottom of D is wider than the top of D, so it constitutes a larger portion of the total weight of D than the top of D. Example 9.7. Find the center of gravity of the domain D whose borders are the vertical lines x =0and x =, the horizontal axis, and the graph of the function f(x) =x 2 = y. Solution: The domain D in question does not have a symmetry line, so we must use Theorem 9.6 to compute both of x g and y g. We have b xf(x) dx a x g = A(D) 0 = x x2 dx 0 x2 dx = [x4 /4] 0 [x 3 /3] 0 = 3 4 and y g = = b a 2 [f(x)]2 dx A(D) 0 (x4 /2) dx /3 = [x5 /0] 0 /3 = 3 0. So the center of gravity of D is at (0.75, 0.3). This agrees with our intuition, since the bottom of D is larger than its top, and the left-hand side of D is smaller than the right-hand side of D. See Figure 9.0 for an illustration Exercises. () Find the center of mass of the unit semicircle that lies in the northern half-plane. (2) Find the center of mass of the plate whose borders are the lines x = 0 and x = π/2, the graph of the function f(x) = sin y, and the horizontal axis.

157 63. APPLICATIONS TO ECONOMICS AND THE LIFE SCIENCES 5 (3) Find the center of mass of the plate whose borders are the vertical lines x = and x = 2, the horizontal axis, and the graph of the function f(x) =lnx. (4) Find the center of mass of the plate whose borders are the vertical lines x = and x = 2, the horizontal axis, and the graph of the function f(x) =e x. (5) Find the center of mass of the trapezoid whose vertices are at (0, 0), (5, 0), (, ), and (8, ). (6) An object consists of two squares. The first is the square with vertices (0, 0), (0, 2), (2, 0), and (2, 2), and the other is the square with vertices (0, 2), (, 2), (0, 3), and (, 3). The density of the material of the small square is twice the density of the material of the large square. Where is the center of mass of this object? 63. Applications to Economics and the Life Sciences 63.. Consumer Surplus. Let us consider the problem of pricing some merchandise whose value is highly subjective; that is, it is worth more to some customers than to others. Examples of this could be tickets for various sporting events, air line tickets to vacation destinations, or popular books. Let p(x) be the demand function of this commodity. That is, p(x) is the price that will result in selling x units of the commodity. Lower prices usually lead to higher sales therefore, p(x) is usually a decreasing function as illustrated in Figure 9.. The area under the graph of p represents the total revenue the company could possibly have, if it managed to charge each customer Figure 9.0. Center of mass of the area defined in Example 9.7.

158 52 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.. The demand function p(x) of a commodity. the maximum price that that customer is willing to pay. Indeed, if the highest amount anyone is willing to pay for one unit is p(x ), and x customers are willing to pay that price, then the revenue coming from these most enthused customers is x p(x ), which is the area of the domain under the graph of p that is between the vertical lines x =0 and y = x. We could continue in this way, noting that if the second highest price that some customers are willing to pay is p(x 2 ), and there are x 2 x people who are willing to pay this price (not including those who are willing to pay even p(x )), then the revenue from them will be (x 2 x )p(x 2 ). This is the area of the domain under the graph of p that is between the lines x = x and x = x 2, and so on. If the seller decides to set one fixed price p(z), then the seller will sell z items, for a total revenue of zp(z) (the area of the rectangle R bordered by the two coordinate axes and the lines x = z and y = p(z)). This means that the customers who would have paid an even higher price for these goods have saved money. Besides losing that potential revenue, the seller also loses revenue by not getting any purchases from customers who were willing to pay some amount, but not z, for one unit. Let x n be the number of items that the seller can sell at the lowest price at which the seller is still willing to sell these items. It is a direct consequence of the above discussion that the total amount saved by all customers who bought the item at z dollars is the area under the curve of p but above the rectangle R, that is, (9.) n (p(x i ) z)(x i x i ). i=

159 63. APPLICATIONS TO ECONOMICS AND THE LIFE SCIENCES 53 If the number n of prices at which various customers are willing to buy goes to infinity, then the Riemann sum in (9.) approaches the definite integral (9.2) CS = z 0 (p(x) z) dx. In economics, CS is called the customer surplus for the given commodity. Similarly, the integral z p(x) dx is the amount of missed revenue, that is, the money the company could have received from buyers who found the product too expensive. Note that this is the area of the domain under the graph of p, but on the right of R. Example 9.8. Tickets for a certain flight are normally priced at $300, and in an average month, 500 tickets are sold. Research shows that, for every $0 that the price is reduced, the number of tickets sold goes up by 20. Find the demand curve and compute the consumer surplus for these tickets if the price is set at $240. Solution: If the airline wants to sell x tickets, then the price that the airline needs to charge is p(x) = x 500 = 300 x Indeed, in order to sell x 500 extra tickets, the airline needs to decrease its price by $0 for each 20 pack of extra tickets. If the price is set at z = $240, then formula (9.2) shows that the customer surplus is CS = = = =60, 000. (p(x) 240) dx ( 60 x ( 30 x 2 ) dx ) dx So customers would save a total of $60,000 in an average month if the price of the tickets were set at $240.

160 54 9. FURTHER APPLICATIONS OF INTEGRATION Survival and Renewal. Example 9.9. Let us assume that there are currently 30,000 people in the United States who have a certain illness. Let us also assume that we know that the fraction of that population who will still have the illness t months from now is given by the function f(t) =e 0.05t. We also know that every month 000 new patients will get the illness. How many people in the United States will have the illness in 20 months? Solution: Clearly, f(20) = e =0.368 of the people who currently have the illness will still have it 20 months from now. Now we have to compute the number of people who will get the illness between now (t = 0) and 20 months from now (t = 20) and will still be ill 20 months from now. Subdivide the interval [0, 20] into n equal subintervals using the points 0=t 0 <t < <t n =20. Set 20/n = t. Then, for all i, there are 000 t people who will get the illness during the time period [t i,t i ). That means that 20 months from now, in other words, approximately 20 t i months from getting the illness, the fraction of them who will still have the illness will be f(20 t i ). So their number will be about 000 t f(20 t i ). Summing over all allowed values of i, we get that the total number of people in the United States who will have the illness 20 months from now is n 30, 000 f(20) t f(20 t i ). i= We recognize that the above sum is a Riemann sum, so, as n goes to infinity, the above expression converges to D =30, 000 f(20) + 20 =30, 000 e , , , f(20 t) dt 0 e 0.05t dt So 20 months from now, 23,679 people in the United States will still have the illness. Note that the result of the previous example shows that the number of people in the United States who have the illness will decrease during

161 63. APPLICATIONS TO ECONOMICS AND THE LIFE SCIENCES 55 the next 20 months. Try to find an intuitive explanation for that fact that does not involve integration Exercises. () A country currently has a population of 80 million people and a natural growth rate of.5 percent. The natural growth g of the population in a given year is computed as the difference between the number of births and the number of deaths in that year, while the natural growth rate for that year is g divided by the size of the population at the beginning of that year. Let us assume that each year. million people emigrate from this country. If the current trends continue, how large will the population of this country be in 20 years? (2) A country currently has a population of 80 million people and a natural growth rate of 0.5 percent. Let us assume that each year 0.35 million people immigrate to this country. If the current trends continue, how large will the population of this country be in 20 years? (3) We deposit $00,000 into a bank account, where it will earn an annual interest of 5 percent. The interest is compounded continuously, so in t years, the original deposit will be worth f(t) = $00, t. Each year, we deposit $2000 to this same account in a continuous manner. What will our account balance be in 5 years? (4) We deposit $00,000 into a bank account, where it will earn an annual interest of 5 percent. The interest is compounded continuously. Each year, we withdraw a total of $4000 in a continuous manner. What will our account balance be in 20 years? (5) Tickets to a certain section of the arena for a basketball game usually cost $50. This results in the sales of 000 tickets. For every dollar that the price is dropped, the number of tickets sold goes up percent. Find the demand function for these tickets and compute the consumer surplus if the tickets are sold at $40. (6) Let S(x) be the supply function for a certain commodity. That is, S(x) is the price that one unit of the commodity has to cost in order to attract enough sellers to provide x units for sale. Note that S(x) is an increasing function, since a higher price is needed to attract more sellers.

162 56 9. FURTHER APPLICATIONS OF INTEGRATION Let us assume that the units are sold at a fixed price T = S(t). That means that the sellers who would be willing to sell at a lower price are making a profit. The total amount of the profit made by all sellers is sometimes called the producer surplus. Prove that the producer surplus for this commodity can be computed by the formula t 0 (T S(x)) dx. 64. Probability The word probability is often used in informal conversations, even if it is sometimes not clear what the speaker means by that word. It turns out that there are two distinct concepts of probability. These two concepts complement each other in that they are applicable in different circumstances, and use very different methods Discrete Probability. Let us say that we are tossing a fair coin four times. What is the probability that we will get at least three heads? This is a situation in which the event that we study, that is, the sequence of four coin tosses, has only a finite number of outcomes. Indeed, there are 2 4 possible outcomes, since each coin toss has two outcomes (heads or tails), and the coin is tossed four times. Among these 6 possible outcomes, five are favorable outcomes, namely, HHHH, HHHT, HHTH, HTHH, and THHH. Furthermore, each single outcome (favorable or not) is equally likely to occur, since the coin is fair, and the result of each coin toss is equally likely to be heads or tails. In this situation, that is, when the number of all possible outcomes is finite, and each outcome is equally likely to occur, define Number of favorable outcomes (9.3) Probability of event =, Number of all outcomes which, in our example, shows that the probability of getting at least three heads is 5/6. Probabilities defined by formula (9.3) are called discrete probabilities. The formula is applicable only when the number of possible outcomes is finite. If we want to apply this formula in complicated situations, we need advanced techniques to count the number of all

163 64. PROBABILITY 57 outcomes and the number of favorable outcomes. The fascinating discipline studying those techniques is called enumerative combinatorics, and will not be discussed in this book Continuous Probability. Let us say we want to know the probability that during the next calendar year the city of Gainesville, Florida, will have more than 40 inches of precipitation, or we want to know the probability that Gainesville will have less than 50 inches of precipitation, or that Gainesville will have at least 42 but at most 48 inches of precipitation. In this case, formula (9.3) is not applicable, since both the number of favorable outcomes and all outcomes is infinite. Indeed, the amount of precipitation in Gainesville next year can be any nonnegative real number. Furthermore, not all outcomes are equally likely. Receiving very little or very much precipitation is far less likely than receiving close to the usual amount. We need a totally different approach. Our approach, while different from the one in the previous section, shares some of the most important features of that approach. For instance, the probability of an event that is certain to happen will be, while the probability of an event that never happens will be 0. Similar situations occur when we want to know the probability that a certain device will work for more than t years, or that a randomly selected person weighs more than p pounds but less than q pounds, or that the blood pressure of a randomly selected person is below a given value. The quantities mentioned here are called random variables. We would like to define the probability F (a) that the amount X of precipitation in Gainesville next year will be at most a inches. This probability will sometimes be denoted by P (X a). While we do not yet know how to compute F (a), we know that the function F has to satisfy the following requirements: () We will have F (a ) F (a 2 )ifa <a 2. In other words, F is increasing. Indeed, if X a, then X a 2 since a <a 2. (2) We will have lim a F (a) =, since the amount of precipitation is always finite. (3) The function F (a) is a continuous function, since a little bit of change in a will only mean a little bit of change in F (a) = P (X a). We sometimes refer to this fact by saying that X is a continuous random variable. Note that the function F is called the distribution of the random variable X.

164 58 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.2. The probability P (a X b) as an area. It can be proved that if F has all these properties, then there exists a unique function f : R R that has the following properties: (a) For all a R, wehave a f(x) dx = F (a) =P (X a). (b) The equality f(x) dx = holds. (c) For all real numbers x, the inequality f(x) 0 holds. If f is the unique function described by the three properties above, then f is called the probability density function, or simply density function, of the continuous random variable X. Note that property (a) above implies that, for all real numbers a<b, the equality (9.4) P (a X b) = b a f(x) dx holds. In other words, P (a X b) is equal to the area of the domain between the graph of the density function f, the horizontal axis, and the vertical lines x = a and x = b. See Figure 9.2 for an illustration. Indeed, we have P (a X b) =P (X b) P (X a) = F (b) F (a) = = b b a f(x) dx f(x) dx. a f(x) dx

165 64. PROBABILITY 59 Figure 9.3. The graph of the function f in Example 9.0. Example 9.0. Let the continuous random variable X have density function 0 if x, f(x) = 6x( x) if 0 x, 0 if x>. Verify that f is indeed a density function and compute the probability P (0.3 X 0.6). Solution: In order to see that f is indeed a density function, we must verify that its definite integral, taken over the entire line of real numbers, is equal to. This is not difficult, since f(x) = 0 outside the interval [0, ]. This leads to f(x) dx =6 0 ( x x 2 ) dx =6[ x 2 2 x3 3 ] 0 =6 6 =. So f is indeed a valid density function. We can use formula (9.4) to compute the requested probability. We get P (0.3 X 0.6) = f(x) dx =6 x x 2 dx 0.3 [ ] x =6 2 x = 6( ) =0.432.

166 60 9. FURTHER APPLICATIONS OF INTEGRATION Figure 9.4. λ = (red), λ =.5 (blue), λ = 2 (orange) Exponential Distribution. Consider the following density function. Let λ be a positive real number and let { 0 if x<0, (9.5) f(x) = λe λx if 0 x. We see that f is a decreasing function on the interval [0, ). Figure 9.4 shows how the speed at which f decreases depends on the parameter λ. It turns out that this density function is a very frequently occurring one. Therefore, it has a name. It is called the exponential density function with parameter λ. Using the right constant λ and under the right circumstances, it can be used in many scenarios, typically connected to waiting times. For instance, it could be used to measure the probability that, given a starting moment, a given cell phone will ring in less than t minutes, or that, at a given location, it will start raining in less than h hours, or that, given a random store, a customer will enter in s seconds. The exponential density function will give a good approximation to compute these probabilities if the mentioned processes take place at a roughly constant rate. That is, we should choose a part of the day when that given cell phone receives calls at roughly constant frequency, a season when it rains at that location at roughly constant time periods, or a time of day when customers enter that store at a roughly constant rate. Example 9.. The probability that a certain kind of new refrigerator will need a major repair in x years is given by the exponential

167 64. PROBABILITY 6 density function with parameter λ =/9. What is the probability that a new refrigerator will not need a major repair for 0 years? Solution: First, we compute the probability that the refrigerator will need a major repair in 0 years. Let X denote the number of years passing before the first major repair is needed. Then that probability is P (X <0) = = f(x) dx 9 e x/9 dx =[ e x/9 ] 0 0 = e 0/9 =0.67. Therefore, the probability that the refrigerator will not need a major repair in 0 years is P (X 0)= 0.67= Mean. If we want to compute the average weight of a person selected from a given population of n people, we can simply take the weights a,a 2,...,a n of those people and compute their arithmetic mean, or average, that is, the real number A = a + a a n. n This could take a very long time if n is a very large number. If the data are given in a more organized form, we may be able to save some time. In particular, if we know that there are b people in the population whose weight is x, there are b 2 people whose weight is x 2, and so on, then we can compute the average weight of the population as (9.6) A = b x + b 2 x b k x k, b + b b k since this fraction is the total weight of the population divided by the number of people in the population. Now note that p i = b i /(b +b 2 + +b k ) is just the probability that a randomly selected person of this population has weight x i. Therefore, (9.6) is equivalent to (9.7) p x + p 2 x p k x k. Theoretically, the weight of a person can take infinitely many values since the measuring scale can be always be more precise. It is not difficult to prove that, as k goes to infinity, the sum in (9.7) will turn

168 62 9. FURTHER APPLICATIONS OF INTEGRATION into a Riemann sum, and the weights x i will be measured by a continuous random variable X, and the probabilities p i will be expressible by the definite integrals of a density function. This leads to the following definition. Definition 9.. Let f be the density function of the continuous random variable X. Then the value of µ(x) = tf(t) dt is called the average value or mean or expected value of X. Example 9.2. Let X be the continuous random variable whose density function is the exponential density function with parameter λ that we defined in (9.5). Then µ(x) =/λ. Solution: Using Definition 9., we have µ(x) = = = tf(t) dt λte λt dt 0 [ te λt ] λ e λt 0 = λ. In other words, the parameter and the mean of an exponential distribution are reciprocals of each other. In view of this, we can reformulate the result of Example 9. as follows. If the average time before a new refrigerator needs a major repair is 9 years, then the probability that a refrigerator will not need a major repair for 0 years is Normal Distribution. Let µ be a real number and let σ be positive real number. Consider the density function f(x) = ( ) (x µ) 2 σ 2π exp. 2σ 2 The distribution defined by this density function is called the normal distribution with parameters µ and σ. This distribution is denoted by N(µ, σ). In particular, if µ = 0 and σ =, then the obtained distribution N(0, ) is called the standard normal distribution. Plotting the graph of f for various values of µ and σ, we see that the graph has a bell curve; its highest point is reached when x = µ,

169 64. PROBABILITY 63 Figure 9.5. σ = (red), σ =.5 (orange), σ = 2 (green), and σ = 2.5 (blue). and it increases on the left of that and decreases on the right of that. The smaller the value of σ, the steeper is the rise and fall of the graph of f. See Figure 9.5 for an illustration. It can be proved that µ is precisely the mean of N(µ, σ). The constant σ is called the standard deviation of N(µ, σ). It measures how spread out the values of our variable X are. (The precise definition is that σ is the square root of the mean of (X µ) 2.) Many scenarios are modeled by a normal distribution, such as test scores, athletic results, or annual snowfall at a given location. Example 9.3. In an average year, Northtown gets 0 feet of snow, with a standard deviation of 2 feet. What is the probability that, in a random year, Northtown gets between 9 and 2 feet of snow if snowfall is modeled by a normal distribution? Solution: Let X denote the snowfall in a random year in Northtown. We need to find the probability P (9 X 2). As snowfall is modeled by a normal distribution, the given parameters imply that that must be the distribution N(0, 2). Therefore, by formula (9.4), we have 2 ( ) (x 0) 2 P (9 X 2) = 9 σ 2π exp 8 =0.5326, where the definite integral has to be computed by some approximation method (or a software package) since e x2 has no antiderivative among elementary functions.

170 64 9. FURTHER APPLICATIONS OF INTEGRATION Exercises. () For which value of c will f(x) =cx 4 be a density function on [0, ]? (2) Let X be a random variable whose density function is 0 outside the interval [0, ] and satisfies f(x) =2x for x [0, ]. Prove that f is indeed a density function, and compute the mean of X. (3) Let us say that the lifetime of a bicycle tire (measured in months) has an exponential distribution with λ = 7. What is the probability that a tire will last between five and eight months? (4) The average score on an exam is 00 points. In order to pass, a student cannot be more than 2 standard deviations below the average. If the scores have a normal distribution with a standard deviation of 6, how large a fraction of the students will pass the exam? (5) Using the conditions of Example 9.3, what is the probability that Northtown will get less than 5 feet of snow in a given year? (6) Let X be the random variable that counts the goals scored by an offensive soccer player of a certain elite league during an entire season. An offensive player is considered exceptional if he the number of goals he scores exceeds the average of all offensive players by at least 3 standard deviations. Let us say that X has distribution N(33, 3). What percentage of offensive players is considered exceptional?

171 CHAPTER 0 Planar Curves 65. Parametric Curves Every point in a plane can be defined as an ordered pair of real numbers (x, y) called the rectangular or Cartesian coordinates. A graph of a function f is the set points in a plane whose coordinates satisfy the condition y = f(x). The graph gives a simple example of a planar curve. More generally, a planar curve can be defined as the set of points whose coordinates satisfy the condition F (x, y) = 0 called the Cartesian equation of a curve. In many instances, an equation F (x, y) =0 has multiple solutions for every given x. For example, consider the circle of unit radius: (0.) x 2 + y 2 = = y = ± x 2, x [, ]. The two solutions represent two semicircles. The graph y = x 2 is the semicircle above the x axis, while the graph y = x 2 is the semicircle below the x axis. The union of the two graphs is the full circle. This example shows a deficiency in describing planar curves by the graph of a function because the curves cannot always be represented as the graph of a single function. On the other hand, (0.) admits a different solution: (0.2) x 2 + y 2 = = x = cos t, y = sin t, t [0, 2π], which immediately follows from the trigonometric identity cos 2 t + sin 2 t = for all values of t. This representation means that a point of the coordinate plane is assigned to every value of t [0, 2π] by the rule (x, y) = (cos t, sin t). The coordinates of points of the circle are functions of a third variable called a parameter. Ast changes, the point (cos t, sin t) traces out the circle of unit radius centered at the origin in the plane. The parameter t has a simple geometrical interpretation. It is the angle counted counterclockwise from the positive x axis to a ray from the origin on which the point (cos t, sin t) lies. This observation admits a natural generalization. Definition 0. (Parametric curves). Let x(t) and y(t) be continuous functions on [a, b]. A parametric curve in the coordinate plane is 65

172 66 0. PLANAR CURVES Figure 0.. Circle: x(t) = cos t, y(t) = sin t. Figure 0.2. Parametric curve. As t increases from a to b, the point (x(t),y(t)) traces out a curve in the xy plane. the set of points satisfying the conditions, called the parametric equations, x = x(t), y = y(t), t [a, b]. The points (x(a),y(a)) and (x(b),y(b)) are called the initial and terminal points of the curve, respectively. The graph of a function f is a particular example of a parametric curve: x = t, y = f(t).

173 65. PARAMETRIC CURVES 67 Figure 0.3. The spiral x = t cos t, y = t sin t, t [0, 2π]. The distance from the origin R = x 2 + y 2 = t increases linearly as the angle t, counted counterclockwise from the positive x axis, increases from 0 to 2π. Parametric curves are common in everyday life. The position of a particle in a plane is defined by its rectangular coordinates (x, y) inthe plane. When the particle moves, its coordinates become functions of time t so that the parametric curve x = x(t), y = y(t) is the trajectory of the particle. The particle moves in a specific direction along its trajectory. The particle may repeat its trajectory (or some portions of it) multiple times. Example 0.. Sketch the curve with the parametric equations x = t cos t, y = t sin t, t [0, 2π]. Solution: A basic approach to visualize the shape of a parametric curve is to plot points (x(t k ),y(t k )), k =, 2,..., n, corresponding to successive values of t: t <t 2 < <t n.forn large enough, a fairly good picture of the curve emerges. This approach can be followed here, and the reader is advised to do so, for example, t k =2πk/n, k =0,,..., n. However, there is another way in part specific to this very problem. Note that x 2 + y 2 = t 2 so that the equations may be written in the form of the parametric equations of the circle x = R cos t, y = R sin t, where the radius increases linearly with t, R = R(t) = t. The parameter t can be viewed as the angle between a ray from

174 68 0. PLANAR CURVES the origin and the positive x axis counted counterclockwise. Thus, the curve has the following interpretation. As the point (x(t),y(t)) rotates about the origin, the distance R between it and the origin increases linearly with the rotation angle. Such a motion occurs along an unwinding spiral. In the interval t [0, 2π], the spiral makes one full turn from the initial point (0, 0) to the terminal point (2π, 0) Parametric Curves and Curves as Point Sets. If a curve is defined as a point set in the coordinate plane, for example, by the Cartesian equation F (x, y) = 0, then there are many parametric equations that describe it. For example, the circle (0.) may also be described by the following parametric equations: (0.3) x 2 + y 2 = = x = cos(3τ), y = sin(3τ), τ [0, 2π]. What is the difference between (0.2) and (0.3)? First, note that, as the parameter t in (0.2) increases, the point (cos t, sin t) traces out the circle counterclockwise (the initial point (, 0) moves upward as y = sin t>0 for 0 <t<π/2). In contrast, the point (cos(3τ), sin(3τ)) does so clockwise with increasing τ (the initial point (, 0) moves downward as y = sin(3τ) for 0 <τ<π/6). Second, as t ranges over the interval [0, 2π], the point (cos t, sin t) traces out the circle only once, while the point (cos(3τ), sin(3τ)) winds about the origin three times because the period of the trigonometric functions involved is 2π/3, so the point returns to the initial point (τ = 0) three times when τ =2π/3, τ =4π/3, and τ =6π/3 =2π. Third, there is a relation between the parameters t and τ: t = 3τ. This example illustrates the main differences between curves defined as a point set by the Cartesian equation F (x, y) = 0 and parametric curves. A parametric curve C is oriented; that is, the point (x(t),y(t)) traces out C in a particular direction (from the initial to the terminal point). A parametric curve may repeat itself multiple times. Parametric equations describing the same point set in the plane differ by the choice of parameter; that is, if (x(t),y(t)) and (X(τ),Y(τ)) trace out the same point set C in the plane, then there is a function g(τ) such that X(τ) =x(g(τ)) and Y (τ) =y(g(τ)). The change of the parameter t = g(τ) is called a reparameterization of a curve C.

175 65. PARAMETRIC CURVES 69 A good mechanical analogy is the motion of a particle. A parametric curve describes the actual motion, that is, how fast and in which direction the particle moves along its trajectory defined as a point set. Example 0.2. Suppose a curve C is described by the parametric equations x = x(t), y = y(t) if t [a, b]. Find the parametric equations of C such that the curve is traced out backward, that is, from the point (x(b),y(b)) to (x(a),y(a)) (the initial and terminal points are swapped). Solution: One has to find a new parameter τ, t = g(τ), such that g(b) =a and g(a) =b. When τ increases from a to b, the parameter t decreases from b to a, and the sought-after parametric equations are obtained by the composition X(τ) =x(g(τ)) and Y (τ) =y(g(τ)). The simplest possibility is to look for a linear relation between t and τ, g(τ) =c + dτ. The coefficients c and d are fixed by the conditions g(a) =b or b = c + da and g(b) =a or a = c + db. Therefore, by subtracting these equations, b a =(c + da) (c + db) = (b a)d or d =. By adding these equations with d =, b+a =(c a)+(c b) or c = a + b. Hence, t =(a + b) τ, so that the parametric equations of C with reversed orientation are x = x(a + b τ), y = y(a + b τ), τ [a, b]. For example, if C is the circle oriented counterclockwise as in (0.2), then the same circle oriented clockwise is described by x = cos(2π τ) = cos τ, y = sin(2π τ) = sin τ, τ [0, 2π] The Cycloid. The curve traced by a fixed point on the circumference of a circle as the circle rolls along a straight line is called a cycloid (see Figure 0.4). To find its parametric equations, suppose that the circle has a radius R and it rolls along the x axis. Let the fixed point P on the circumference be initially at the origin so that the center of the circle is positioned at the point (0,R) (on the y axis). Let CP denote the straight line segment between the center of the circle C and P. Initially, CP is perpendicular to the x axis. As the circle rolls, the segment CP rotates about the center of the circle. Therefore, it is natural to choose the angle of rotation θ as a parameter. The coordinates of P are functions of θ to be found. If the circle rolls a distance D so that its center is at (D, R), then the arc length Rθ of the part of the circle between P and the touch point T has to be equal to D, that is, D = Rθ. Let Q be a point on the segment CT such that PQ and CT are perpendicular. Consider the right-angled triangle CPQ. Its hypotenuse CP has length CP = R, and the lengths of its catheti are CQ = CP cos θ = R cos θ and PQ = CP sin θ = R sin θ. Let

176 70 0. PLANAR CURVES Figure 0.4. Definition of a cycloid. A disk of radius R is rolling along the x axis. A curve traced out by a fixed point on its edge is called a cycloid. Figure 0.5. Overall shape of a cycloid. (x, y) be coordinates of the point P. The parametric equations of the cycloid are x = D PQ = Rθ R sin θ = R(θ sin θ). y = R CQ = R R cos θ = R( cos θ). It looks like an upward arc over the interval 0 x 2πR, with maximal height y max =2R (θ = π/2), and the arc repeats itself over the next interval of the length of circumference 2πR and so on. Remark. In 696, the Swiss mathematician Johann Bernoulli posed the brachistochrone problem: Find the curve along which a particle will slide (without friction) in the shortest time (under the influence of gravity) from a point A toalowerpointb not directly beneath A. The particle will take the least time sliding from A to B if the curve is a part of an inverted arch of a cycloid.

177 65. PARAMETRIC CURVES Families of Curves. Different values of R define different cycloids. In general, if the parametric equations contain a numerical parameter, then the parametric equations define a family of curves; each family member corresponds to a particular value of the numerical parameter. Example 0.3. Investigate the family of curves with the parametric equations x = a cos t, y = b sin(2t), t [0, 2π], where a and b are positive numbers. Solution: Consider first the simplest case a = b =. The function x(t) = cos t has a period of 2π, and y(t) = sin(2t) has a period of π. The initial point is (x(0), y(0)) = (, 0). As t increases, the point moves upward so that x(t) decreases (becomes less than ), while y(t) increases, reaching its maximum value at t = π/4. After that, y(t) begins to decrease, while x(t) continues to decrease. At t = π/2, the point arrives at the origin and passes through it into the third quadrant so that x(t) and y(t) continue to decrease. When t =3π/4, y(t) attains the minimum value and begins to increase for t > 3π/4, while x(t) = cos t is still decreasing toward its minimal value, which is reached at t = π, and the curve crosses the x axis moving into the second quadrant. In the second quadrant π<t<3π/2, x(t) increases toward 0, while y(t) first reaches its maximum value at t = π + π/4 (the curve touches the horizontal line y = ) and then decreases to 0. The curve passes the origin again at t =3π/2 and moves into the fourth quadrant, where it again touches the horizontal line y = at t =3π/2+π/4 and at t =2π it arrives at the initial point. The shape of the curve resembles the infinity sign ( ) embedded into the square bounded by the lines y = ± and x = ± so that it touches each of the horizontal sides y = ± twice and each of the vertical sides x = ± once. If a and b are arbitrary, the transformation x ax stretches (a >) or compresses (a <) any geometrical set horizontally in the coordinate plane. The transformation y by does the same but in the vertical direction. So the family of curves consists of curves of the shape stretched to fit into the rectangle bounded by the lines x = ±a and y = ±b Exercises. In () (9), sketch the curve by plotting its points. Include the arrow showing the orientation of the curve. Eliminate the parameter to find a Cartesian equation of the curve.

178 72 0. PLANAR CURVES () x =+2t, y =3 t (2) x = t, y =2 t (3) x = t 2,y= t 3 (4) x =+2e t,y=3 e t (5) x = cosh t, y = sinh t (6) x = 2 sin t, y = 3 cos t (7) x =2 3cos t, y = + sin t (8) x = cos t, y = sin(4t) (9) x = t 2 sin t, y = t 2 cos t (0) The curves x = a sin(nt), y = b cos t, where n is a positive integer, are called Lissajous figures. Investigate how these curves depend on a, b, and n. () Consider a disk of radius R. Let P be a point on the disk at a distance b from its center. Find the parametric equations of the curve traced out by the point P as the disk rolls along a straight line. The curve is called a trochoid. Are the equations well defined if b>r. Sketch the curve for b<r, b = R, and b>r. (2) The swallowtail catastrophe curves are defined by the parametric equations x =2ct 4t 3, y = ct 2 +3t 4. Sketch these curves for a few values of c. What features do the curves have in common? How do they change when c increases? 66. Calculus with Parametric Curves 66.. Tangent Line to a Parametric Curve. Consider a parametric curve x = x(t), y = y(t), where the functions x(t) and y(t) are continuously differentiable and the derivatives x (t) and y (t) do not vanish simultaneously for any t. Such parametric curves are called smooth. Theorem 0. (Tangent Line to a Smooth Curve). A smooth parametric curve x = x(t), y = y(t) has a tangent line at any point (x 0,y 0 ), and its equation is (0.4) x (t 0 )(y y 0 ) y (t 0 )(x x 0 )=0, where (x 0,y 0 )=(x(t 0 ),y(t 0 )). Proof. Take a point of the curve (x 0,y 0 )=(x(t 0 ),y(t 0 )) corresponding to a particular value t = t 0. Suppose that x (t 0 ) 0. Then, by the continuity of x (t), there is a neighborhood I δ =(t 0 δ, t 0 + δ) for some δ > 0 such that x (t) 0 for all t I δ ; that is, the derivative is either positive or negative in I δ. By the inverse function theorem (studied in Calculus I), there is an inverse function t = f(x) that is differentiable in some open interval that contains x 0. Substituting t = f(x) into the second parametric equation y = y(t), one obtains that near the point (x 0,y 0 ) the curve can be represented as a part of the graph y = F (x) such that y 0 = F (x 0 ). The function F is

179 66. CALCULUS WITH PARAMETRIC CURVES 73 differentiable as the composition of two differentiable functions. The derivative F (x 0 ) determines the slope of the tangent to the graph, and the equation of the tangent line reads (0.5) y = y 0 + F (x 0 )(x x 0 ). By construction, y = F (x) = y(t) =F (x(t)) for all t I δ. Differentiation of this equation with respect to t by means of the chain rule yields y (t) =F (x(t))x (t) = F (x(t)) = y (t) x (t) = F (x 0 )= y (t 0 ) x (t 0 ). Substituting this equation into (0.5), the latter can be written in the form (0.4). If x (t 0 ) = 0, then y (t 0 ) 0 by the definition of a smooth curve so that there is a differentiable inverse t = g(y) and hence x = G(y) = x(g(y)). Similar arguments lead to the conclusion that the tangent line to the graph x = G(y) has the form (0.4). The details are left to the reader as an exercise. The rule for calculating the slope of the tangent line can also be obtained by means of the concept of the differential. Recall that the differentials of two related quantities y = F (x) are proportional: dy = F (x) dx. On the other hand, x = x(t), y = y(t) and therefore dx = x (t) dt and dy = y (t) dt. Hence, F (x) = dy dy dx = dt dx dt = y (t) x (t). These manipulations with differentials are based on a tacit assumption that, for a smooth curve x = x(t), y = y(t), there exists a differentiable function F such that y = F (x). In the proof of the tangent line theorem, this has been shown to be true as a consequence of the inverse function theorem. The use of the differentials establishes the following helpful rules to calculate the derivatives: d d dx = dt dx dt = d x (t) dt and d d dy = dt = d y (t) dt Concavity of a Parametric Curve. The concavity of a graph y = F (x) is determined by the sign of the second derivative F (x). If F (x) > 0, the graph is concave downward, and it is concave upward if dy dt

180 74 0. PLANAR CURVES F (x) < 0. If y(t) and x(t) are twice differentiable, then the concavity of the curve can be determined: ( ) d 2 y d 2 x = d dy dx dx = y d ( dy ) x = = y x x y. x dt dx x (x ) 3 Example 0.4. A curve C is defined by the parametric equations x = t 2, y = t 3 3t. (i) Show that C has two tangent lines at the point (3, 0). (ii) Find the points on C where the tangent line is horizontal or vertical. (iii) Determine where the curve is concave upward or downward. Solution: (i) Note that y(t) =t(t 2 3) = 0 has three solutions t =0 and t = ± 3. But the curve has only two points of intersection with the x axis, (0, 0) and (3, 0), because x(± 3) = 3; that is, the curve is self-intersecting at the point (3, 0). This explains why the curve may have two tangent lines. One has x (t) =2t and y (t) =3t 2 3 so that x (± 3) = ±2 3 and y (± 3) = 6. So the slopes of the tangent lines are (y /x )(± 3) = ± 3, and the equations of the lines read y = 3(x 3) and y = 3(x 3). (ii) The tangent line becomes horizontal when y (t) =3t 2 3=0 (see Eq. (0.4)). This happens when t = ±. Thus, the tangent line is horizontal at the points (, ±2). The tangent line is vertical if x (t) =2t =0ort = 0. So the tangent line is vertical at the origin (0, 0). (iii) The second derivative is d 2 y d 2 x = d dy x dt dx = d 3t 2 3 = 3 d ( t ) = 3 (+ ). 2t dt 2t 4t dt t 4t t 2 This equation shows that the curve is concave downward if t>0 (the second derivative is positive) and the curve is concave upward if t<0 (the second derivative is negative) Cusps of Planar Curves. Consider a curve defined by the Cartesian equation x 2 y 3 = 0. This equation can be solved for y, y = x 2/3, such that dy/dx = 2 3 x /3. For x>0, the slope of the tangent line diverges, y (x) as x 0 + (as x approaches to 0 from the right). For x<0, it also diverges, y (x) as x 0 (as x approaches 0 from the left). The two branches of the curve (x >0 and x<0) are joined at x = 0 and have a common tangent line, which is the vertical line x = 0 (the y axis) in this case, but the slope suffers a jump discontinuity (from to ). So the curve is not smooth at x = 0 and

181 66. CALCULUS WITH PARAMETRIC CURVES 75 Figure 0.6. Plot of y = x 2/3. The curve has a cusp at the origin. exhibits a horn like shape near x = 0. Such a point of a planar curve is called a cusp. A parametric curve x = x(t), y = y(t) may have cusps even though both derivatives x (t) and y (t) are continuous for all t. For example, consider the parametric curve x = t 3, y = t 2. For all values of t, x 2 y 3 = 0. So this curve coincides with that discussed above and has a cusp at the origin (t = 0). The derivatives x (t) =3t 2 and y (t) =2t are continuous everywhere, and, in particular, x (0) = y (0) = 0 at the cusp point. Despite the continuity of the derivatives, the slope of the curve is not defined since dy/dx = y /x is an undetermined form 0. 0 A closer investigation shows that the slope y (t)/x (t) = 2 3 t suffers a jump discontinuity (from to + as t changes from negative to positive). The definition of a smooth parametric curve requires that the derivatives x (t) and y (t) are continuous and do not vanish simultaneously at any t. A rationale for the latter condition is to eliminate possible cusps that may occur at points where both derivatives vanish. Furthermore, consider the curve x = t 2, y = t 3. The slope dy/dx = y /x = 3 t is continuous everywhere and, in particular, at t = 0, where 2 x (0) = y (0) = 0. Nevertheless, the curve has a cusp at the origin. To see this, let us investigate the Cartesian equation of this curve x 3 y 2 = 0, which can be solved for x, x = y 2/3. Therefore, the derivative dx/dy = 2 3 y /3 exhibits a jump discontinuity from to as y changes from negative to positive. The two branches of the curve (y >0 and y<0) have a common tangent line (the horizontal line y = 0), but at their joining point a cusp is formed. Note also that the rate dx/dy = x /y = 2 3 t suffers a familiar infinite jump discontinuity, thus indicating a cusp. This example shows that both rates dy/dx = y /x

182 76 0. PLANAR CURVES Figure 0.7. The curve x = t 2, y = t 3. The derivatives x (0) = y (0) = 0 vanish at t = 0. The curve has a cusp at the point (x(0),y(0)) = (0, 0). and dx/dy = x /y must be studied to determine whether there is a cusp at the point where y = x =0. Example 0.5. Find the tangent line to the astroid defined by the parametric equation x = a cos 3 t, y = a sin 3 t, t [0, 2π] at the points t = π/4. Determine the points where the tangent line is horizontal and vertical. Is the curve smooth? Specify the regions of upward and downward concavity. Use the results to sketch the curve. Solution: The slope of the tangent line at a generic point is dy dx = d x dt y = 3a cos 2 t sin t 3a sin2 t cos t = tan t. The value t = π/4 corresponds to the point x = a/2 3/2, y = a/2 3/2 because sin(π/4) = cos(π/4)=/ 2, and the slope at this point is. So the tangent line is y = a ( 2 2 x a ) 2 or y = a x. 2 2 The slope dy/dt = tan t vanishes at t = 0 and t = π so the tangent line is horizontal (y = 0) at the points (±a, 0). However, the derivatives x (t) = 3a cos 2 t sin t and y (t) =3asin 2 t cos t vanish simultaneously

183 66. CALCULUS WITH PARAMETRIC CURVES 77 at t = 0 and t = π. The inverse slope dx/dy =/(dy/dx) = cot t exhibits an infinite jump discontinuity at t = 0 and t = π, so the curve has cusps at (±a, 0) and hence it is not smooth at these points. The slope dy/dx is infinite at t = π/2 and t =3π/2. Therefore, the curve has a vertical tangent line (x =0)at(0, ±a). However, the slope dy/dx = tan t has an infinite jump discontinuity at t = π/2 and t =3π/2. So the curve has cusps and is not smooth at (0, ±a). Note also that both derivatives x and y vanish at these points. Thus, the curve consists of four smooth pieces, and the curve has cusps at the joining points of its smooth pieces. The second derivative d 2 y d 2 x = x d dt dy dx = 3a cos 2 t sin t (tan t) = 3a sin t cos 4 t is positive if sin t>0 (or y>0) and negative if sin t<0 (or y<0). So the two branches of the curve above the x axis are concave downward, while the two branches below it are concave upward. The curves look like a square with vertices (±a, 0), (0, ±a) whose sides are bent inward toward the origin Exercises. In () and (2), find an equation of the tangent line(s) to the curve at the given point. Sketch the curve and the tangent(s). () x = t 2 + t, y = 4 sin t, (0, 0) (2) x = sin t + sin(2t),y= cos t + cos(2t), (, ) In (3) (6), investigate the concavity of the curve. (3) x = t 3 2t, y = t 2 (4) x = sin(2t),y= cos t (5) x = t 2 ln t, y = t 2 +lnt (6) x = 3 sin t 3,y= 2 cos t 3 (7) Investigate the slope of the trochoid x = Rφ b sin φ, y = R b cos φ in terms of φ. Find the condition on the parameters R and b such that the trochoid has vertical tangent lines. (8) At what points on the curve x =2t 3, y = +4t t 2 does the tangent line have slope? (9) Find equations of the tangents to the curve x =2t 3 +, y =3t 2 + that pass through the point (3, 4). In (0) (2). Investigate whether the curve has cusps or not. If it does, find their position. Sketch the curve. (0) x = t 3,y= t 3 () x = t 5,y= t 2 (2) x =(t 2 ) 3,y=(t 3 ) 2

184 78 0. PLANAR CURVES 67. Polar Coordinates A point on a plane is described by an ordered pair of numbers (x 0,y 0 ) in the rectangular coordinate system. This description implies a geometrical procedure to obtain the point as the intersection of two mutually perpendicular lines x = x 0 and y = y 0. The set of vertical and horizontal lines form a rectangular grid in a plane. There are other possibilities to label points on a plane by ordered pair of numbers. Here the polar coordinate system is introduced, which is more convenient for many purposes. Fix a point O on a plane. Let P be a point on the plane. A horizontal ray from O is called the polar axis, and the point O is called the origin or pole. Let θ be the angle between the polar axis and the ray OP from O through P. The angle θ is counted counterclockwise from the polar axis. The position of the point P on the ray OP is uniquely determined by the distance r = OP. Thus, any point P on a plane is uniquely associated with the ordered pair (r, θ), and r, θ are called the polar coordinates of P. The coordinate r is called the radial variable, and θ is called the polar angle. All points on the plane that have the same value of the radial variable form a circle of radius r centered at the origin (all points that have the same distance from the origin). All points on the plane that have the same value of the polar angle form a ray (a half-line bounded by the Figure 0.8. Definition of the polar coordinates in a plane. r is the distance OP, and θ is the angle counted counterclockwise from the horizontal ray outgoing from O to the right. The rectangular coordinates of a point P are related to the polar ones as x = r cos θ, y = r sin θ.

185 67. POLAR COORDINATES 79 origin). So a point P with polar coordinates (r, θ) is the intersection of the circle of radius r and the ray that makes the angle θ with the polar axis. Concentric circles and rays originating from the center of the circles form a polar grid in a plane (see Figure 0.9). To represent all points of a plane, the radial variable has to range over the interval r [0, ), while the polar angle takes its values in the interval [0, 2π) because any ray from the origin does not change after rotation about the origin through the angle 2π. It is convenient, though, to let θ range over the whole real line. Positive values of θ correspond to rotation angles counted counterclockwise, while negative values of θ are associated with rotation angles counted clockwise. All pairs (r, θ) with a fixed value of r and values of θ different by integer multiples of 2π represent the same points of the plane. For example, the ordered pairs (r, θ) =(, π) and (,π) correspond to the same point. Indeed, both points are on the circle of unit radius. The ray θ = π is obtained from the polar axis by counterclockwise rotation of the latter through the angle π. But the same ray is obtained by rotating the polar axis through the angle π clockwise; that is, the rays θ = π and θ = π coincide. Figure 0.9. Polar grid. Coordinate curves of the polar coordinates. The curves of constant values of r are concentric circles. The curves of constant values of θ are rays outgoing from the origin.

186 80 0. PLANAR CURVES Furthermore, the meaning of the radial variable r can be extended to the case in which r is negative by agreeing that the pairs ( r, θ) and (r, θ + π), r > 0, represent the same point. Geometrically, the points (±r, θ) lie on a line through the origin at the same distance r from the origin but on the opposite sides of the origin. With this agreement on extending the meaning of the polar coordinates, each point on a plane may be represented by countably many pairs: (0.6) (r, θ) (r, θ +2πn) or ( r, θ +(2n +)π), where n is an integer Rectangular and Polar Coordinates. Suppose that the polar axis is set so that it coincides with the positive x axis of the rectangular coordinate system. Every point on the plane is either described by the rectangular coordinates (x, y) or the polar coordinates (r, θ). It is easy to find the relation between the polar and rectangular coordinates of a point P by examining the rectangle with the diagonal OP. Its horizontal and vertical sides have lengths x and y, respectively. The length of the diagonal is r. The angle between the horizontal side and the diagonal is θ. Therefore, cos θ = x/r and sin θ = y/r, or x = r cos θ, y= r sin θ r 2 = x 2 + y 2, tan θ = y x. These relations allow us to convert the polar coordinates of a point to rectangular coordinates and vice versa. Example 0.6. Find the rectangular coordinates of a point whose polar coordinates are (2,π/6). Find the polar coordinates of a point with rectangular coordinates (, ). Solution: For r = 2 and θ = π/6, one has x = 2 cos(π/6)=2 3/2 = 3 and y = 2 sin(π/6)=2/2 =,so(x, y) =( 3, ). For x = and y =, one has r 2 =2orr = 2 and tan θ =. The point (, ) lies in the second quadrant, that is, π/2 θ π. Therefore, θ =3π/4. Alternatively, one can take θ =3π/4 2π = 5π/ Polar Graphs. A polar graph is a curve defined by the equation r = f(θ) or, more generally, F (r, θ) = 0. It consists of all points that have at least one polar representation (r, θ) that satisfies the equation. Here polar coordinates are understood in the extended sense of (0.6) when they are allowed to take any value. The simplest polar graph is defined by a constant function r = a, where a is real. Since r represents the distance from the origin, the

187 67. POLAR COORDINATES 8 pairs ( a,θ) form a circle of radius a centered at the origin. Similarly, the graph θ = b, where b is real, is the set of all points (r, b), where r ranges over the real axis, which is the line through the origin that makes an angle b radians with the polar axis. Notice that the points (r, b), r>0, and (r, b), r<0, lie in the opposite quadrants relative to the origin as the pairs (r, b) and ( r, b + π) represent the same point. In general, the shape of a polar graph can be determined by plotting points (f(θ k ),θ k ), k =, 2,..., n, for a set of successively increasing values of θ, θ <θ 2 < <θ n ; that is, one takes a set of rays θ = θ k and marks the point on each ray at a distance r k = f(θ k ) from the origin. Example 0.7. Describe the curve r = 2 cos θ. Solution: By converting the polar graph equation to rectangular coordinates, one finds: r = 2 cos θ r 2 =2rcos θ x 2 + y 2 =2x (x ) 2 + y 2 =. The latter equation is obtained by completing the squares. It represents a circle with center (, 0) and radius. Note also that by looking at the graph of the cosine function, one can see that the point (2 cos θ, θ) gets closer to the origin when θ changes from 0 to π/2 (the first quadrant), reaching the origin at θ = π/2. This gives the upper part of the circle. A similar behavior is observed when θ changes from 0 to π/2 (the lower part of the circle in the fourth quadrant). In the intervals ( π, π/2) and (π/2,π), the radial variable is negative. The representation (r, θ) is equivalent to ( r, θ ± π). Therefore, the points (2 cos θ, θ) and ( 2 cos θ, θ + π) = (2 cos(θ + π),θ + π) are the same for θ [ π, π/2]. But the latter set can also be described by the pairs (2 cos θ, θ) ifθ [0,π/2]. Similarly, the set traced out by the pair (2 cos θ, θ) for θ [π/2,π] is the same as when θ [ π/2, 0]. So the pair (2 cos θ, θ) traces out the same set (the circle) each time θ ranges an interval of length π. Example 0.8. Describe the shape of the curve r = θ, θ 0. Solution: The point (θ, θ) lies on the ray that makes an angle θ with the polar axis and is a distance r = θ from the origin. As the ray rotates counterclockwise about the origin with increasing θ, the distance of the point from the origin increases proportionally. So the curve is a spiral unwinding counterclockwise Symmetry of Polar Graphs. When sketching polar graphs, it it is helpful to take advantage of symmetry, just like when plotting graphs

188 82 0. PLANAR CURVES Figure 0.0. Polar curve r = θ. It is a spiral because the distance from the origin r increases with the angle θ as the point rotates about the origin through the angle θ. y = f(x) for symmetric (f( x) =f(x)) or skew-symmetric (f( x) = f(x)) functions. (i) If a polar equation is unchanged when θ is replaced by θ, the curve is symmetric about the polar axis. Note that the transformation (r, θ) (r, θ) means that (x, y) (x, y), which is the reflection about the x axis (or the polar axis). (ii) If a polar equation is unchanged when (r, θ) is replaced by ( r, θ) orby(r, θ + π), the curve is symmetric about the origin. Again, these transformations are equivalent to (x, y) ( x, y), which is the reflection about the origin. (iii) If the equation is unchanged under the transformation (r, θ) (r, π θ), then the curve is symmetric about the vertical line θ = π/2. In the rectangular coordinates, this transformation is (x, y) (x, y), which is the reflection about the y axis. Example 0.9. Describe the cardioid r = + sin θ. Solution: The equation is unchanged under θ π θ so the curve is symmetric about the vertical axis (the y axis). It is sufficient to investigate the curve in the interval θ [ π/2,π/2] (in the fourth and first quadrants). Consider a ray that rotates counterclockwise from θ = π/2 toθ = π/2 (from the negative y axis to the positive y axis). When θ = π/2, r =0. Asθ increases from π/2 to 0 (the fourth quadrant), the distance from the origin r = + sin θ increases

189 67. POLAR COORDINATES 83 Figure 0.. The cardioid r = + sin θ. monotonically from 0 to (r = on the polar axis). In the interval [0,π/2] (the first quadrant), the distance from the origin r continues to increase monotonically and reaches its maximal value 2 on the vertical axis Tangent to a Polar Graph. To find a tangent line to a polar graph r = f(θ), the polar angle is viewed as a parameter so that the parametric equations of the graph are x = r cos θ = f(θ) cos θ, y= r sin θ = f(θ) sin θ. By the product rule for the derivative, dy dy dx = dθ dx dθ = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ. In particular, if the curve passes through the origin, r = 0, the equation for the slope at the origin is simplified dy dx = tan θ if dr dθ = f (θ) 0. Note that if f (θ) = 0, then the slope is an undetermined form 0 0 because x (θ) =y (θ) = 0 for any value of θ such that f (θ) =f(θ) =0. This means that the curve may have a cusp at the origin and hence is not smooth. Example 0.0. Find the slope of the cardioid r = + sin θ in terms of θ. Investigate the behavior of the cardioid near the origin.

190 84 0. PLANAR CURVES Solution: Here f(θ) = + sin θ and f (θ) = cos θ. This leads to the slope dy dx = cos θ sin θ +(+sinθ) cos θ cos 2 θ ( + sin θ) sin θ = cos θ( + 2 sin θ) ( + sin θ)( 2 sin θ). where the identity cos 2 θ = sin 2 θ has been used to transform the denominator. The cardioid passes through the origin as θ passes through π/2. The slope dy/dx is undetermined because the numerator and denominator of the ratio vanish at θ = π/2 (both derivatives dx/dθ and dy/dθ vanish). The left and right limits have to be investigated to see if the slope has a jump discontinuity thus indicating a cusp. The numerator vanishes because of the factor cos θ, while the denominator vanishes because of the factor ( + sin θ). Hence, dy lim θ ( π/2) ± dx = 3 lim θ ( π/2) ± cos θ + sin θ = 3 sin θ lim θ ( π/2) ± cos θ =, where l Hospital s rule has been used to resolve the undetermined form 0 and the property that tan θ as θ 0 ( π/2)± has been invoked to find the limit. The cardioid has a vertical tangent line at the origin. The slope has an infinite jump discontinuity, meaning that the cardioid has a cusp at the origin (see Figure 0.) Exercises. In () (3), convert the polar graph equation to a Cartesian equation and sketch the curve. () r = 4 sin θ (2) r = tan θ sec θ (3) r = 2 sin θ 4 cos θ In (4) (2), sketch the curve with the given polar equation. (4) r = θ, θ 0 (5) r =lnθ, θ (6) r 2 3r +2=0 (7) r = 4 cos(6θ) (8) r 2 = 9 sin(2θ) (9) r = + 2 cos(2θ) (0) r = 2 + sin(3θ) () r = + 2 sin(3θ) (2) r 2 θ = (3) Sketch the curve (x 2 + y 2 ) 2 =4x 2 y 2. Hint: Use polar coordinates. (4) Investigate the dependence of the shape of the curve r = cos(nθ) as the integer n increases. What happens if n is not an integer? (5) Show that the curve r =+a sin θ has an inner loop when a > and find the range of θ that corresponds to the inner loop. (6) For what values of a is the curve r =+a sin θ smooth? In (7) and (8), find the slope of the tangent line to the given curve at the point specified by the value of θ and give an equation of the

191 68. PARAMETRIC CURVES: THE ARC LENGTH AND SURFACE AREA 85 tangent line. (7) r = 2 sin θ, θ= π/3 (8) r = 2 cos θ, θ= π/6 (9) Show that the curves r = a sin θ and r = a cos θ intersect at right angles. 68. Parametric Curves: The Arc Length and Surface Area 68.. Arc Length of a Smooth Curve. Let C be a smooth curve defined by the parametric equations x = x(t), y = y(t), where t [a, b]. Suppose that C is traversed exactly once as t increases from a to b and consider a partition of the interval [a, b] such that t 0 = a and t k = t 0 + k t, k = 0,, 2,..., n, are the endpoints of the partition intervals of width t =(b a)/n. Then the points P k with coordinates (x(t k ),y(t k )) lie on the curve so that P 0 and P n are the initial and terminal points, respectively. The curve C can be approximated by a polygonal path with vertices P k. By definition, the length L of C is the limit of the lengths of these approximating polygons as n : (0.7) L = lim n n P k P k, k= provided the limit exists, and, in this case, the curve is called measurable. Figure 0.2. The arc length of a smooth parametric curve is approximated by the length of n straight line segments connecting points on the curve. The arc length is defined in (0.7) as the limit n.

192 86 0. PLANAR CURVES By the mean value theorem, when applied to the functions x(t) and y(t) on the interval [t k,t k ], there are numbers t k and t k in (t k,t k ) such that x k = x(t k ) x(t k )=x (t k) t, y k = y(t k ) y(t k )=y (t k ) t. Therefore, P k P k = ( x k ) 2 +( y k ) 2 = (x (t k ))2 +(y (t k ))2 t. The sum in (0.7) resembles a Riemann sum for the function F (t) = (x (t)) 2 +(y (t)) 2. It is not exactly a Riemann sum because t k t k in general. However, if x (t) and y (t) are continuous, it can be shown that the limit (0.7) is the same as if t k and t k were equal, namely, L is the integral of F (t) over [a, b]. Theorem 0.2 (Arc Length of a Curve). If a curve C is described by the parametric equations x = x(t), y = y(t), t [a, b], where x (t) and y (t) are continuous on [a, b] and C is traversed exactly once as t increases from a to b, then the length of C is b (dx ) 2 ( dy ) 2 L = + dt. a dt dt If C is a graph y = f(x), then x = t, y = f(t), and dx = dt, and the length is given by the familiar expression b ( dy ) 2 L = + dt. a dx It is convenient to introduce the arc length of an infinitesimal segment of a curve (the differential of the arc length) ds = b ds (dx) 2 +(dy) 2 = L = ds = C a dt dt. The symbol means the summation over infinitesimal segments of the C curve S (the integral along a curve C) and expresses a simple fact that the total length is the sum of the lengths of its (infinitesimal) pieces Independence of Parameterization. By its very definition, the arc length is independent of the parameterization of the curve. If a curve C is defined as a point set, then any parametric equations can be used to evaluate the arc length. Let C be traced out only once by x = x(t), y = y(t), where t [a, b], and by x = X(τ), y = Y (τ), where τ [α, β]. As noted, there is a relation between the parameters t and τ, τ = g(t), such that g(t) increases from α to β as t increases from a to b, that

193 68. PARAMETRIC CURVES: THE ARC LENGTH AND SURFACE AREA 87 dτ is, = g (t) 0, such that x(t) = X(g(t)) and y(t) = Y (g(t)). dt Therefore, the integrals (0.7) corresponding to different parametric equations of the same curve are related by a change of the integration variable: b (dx ) 2 ( dy ) 2 b (dx dτ ) 2 ( dy dτ ) 2 L = + dt = + dt a dt dt a dτ dt dτ dt b (dx ) 2 ( dy ) 2 dτ β (dx ) 2 ( dy ) 2 = + a dτ dτ dt dt = + dτ. α dτ dτ Thus, the arc length is independent of the curve parameterization and can be computed in any suitable parameterization of the curve. A circle of radius R is described by the parametric equations x = R cos t, y = R sin t, t [0, 2π]. Then dx = R sin tdt and dy = R cos tdt. Hence, ds 2 =(Rsin tdt) 2 +(R cos tdt) 2 = R 2 (sin 2 t+cos 2 t) dt 2 = R 2 dt 2,ords = Rdt, and 2π 2π L = ds = Rdt= R dt =2πR. C 0 0 Example 0.. Find the length of one arch of the cycloid x = R(φ sin φ), y = R( cos φ). Solution: According to the description of the cycloid, one arch corresponds to the interval φ [0, 2π]. The arc length differential ds is found as follows: dx = R( cos φ) dφ, dy = R sin φ dφ, ds 2 = dx 2 + dy 2 = [( cos φ) 2 + sin 2 φ]r 2 dφ 2 =[ 2 cos φ + cos 2 φ + sin 2 φ]r 2 dφ 2 =(2 2 cos φ)r 2 dφ 2. ds = 2( cos φ) R dφ. To evaluate the integral of 2( cos φ), the double-angle identity is invoked, sin 2 (φ/2)=( cos φ)/2. Since 0 φ/2 π when φ [0, 2π], the sinus is nonnegative, sin(φ/2) 0, in the integration interval, and hence, after taking the square root ( u 2 = u ), the absolute value can be omitted. Thus, 2π 2π L = R 4 sin 2 (φ/2) dφ =2R sin(φ/2) dφ 0 0 = 2R[ 2 cos(φ/2)] =8R. 2π 0

194 88 0. PLANAR CURVES Area of a Planar Region. The area under the curve y = f(x) and above the interval x [a, b] is given by A = b f(x) dx, where a f(x) 0. Suppose that the curve is also described by parametric equations x = x(t), y = y(t), so that the function x(t) is one-to-one. Then, by changing the integration variable, dx = x (t) dt and A = b a ydx= β α y(t)x (t) dt. The new integration limits are found as usual. When x = a, t is either α or β, and when x = b, t is the remaining value. Example 0.2. Find the area under one arch of the cycloid x = R(φ sin φ), y = R( cos φ). Solution: When φ [0, 2π], x [0, 2πR] for one arch of the cycloid, and y(φ) 0. Using the differential dx found in the previous example, 2πR 2π A = ydx= R 2 ( cos φ) 2 dφ 0 2π = R 2 ( 2 cos φ + cos 2 φ) dφ 0 0 2π 2π = R 2 [ + ( + cos(2φ))] dφ = 2 R2 ( + ) dφ 2 =3πR2, 0 where 2π cos φdφ= 0 and 2π cos(2φ) dφ = 0 by the 2π periodicity of 0 0 the cosine function Surface Area of Axially Symmetric Surfaces. An axially symmetric surface is a surface symmetric relative to rotations about a line. Such a line is called the symmetry axis. For example, a cylinder is symmetric relative to rotations about its axis, a sphere is symmetric relative to rotations about its diameter, and so on. An axially symmetric surface is swept by a planar curve when the latter is rotated about a line. A cylinder of radius R and height h is obtained by revolving a straight line segment of length h about a line parallel to the segment at a distance R. A sphere of radius R is obtained by revolving a circle of radius R about its diameter. Let ds be the arc length of an infinitesimal segment of a smooth curve C positioned at a point (x, y). If the distance between the point (x, y) and the symmetry axis is R(x, y), then the area da of the part of the surface swept by the curve segment when the latter is rotated about the symmetry axis is the area of a cylinder of radius R(x, y) and 0

195 68. PARAMETRIC CURVES: THE ARC LENGTH AND SURFACE AREA 89 height ds: da =2πR(x, y) ds. The total surface area is the sum of areas of all such parts of the surface (0.8) A =2π R(x, y) ds =2π C b a R(x(t),y(t)) ds dt dt, where x = x(t), y = y(t), a t b are parametric equations of C. Here it is again assumed that the point (x(t),y(t)) traces out the curve C only once as t increases from a to b. In particular, if the symmetry axis coincides with the x axis, then R(x, y) = y (the distance of the Figure 0.3. A surface is obtained by rotation of a smooth curve about a vertical line. If ds is the arc length of an infinitesimal segment of the curve at a point P and R is the distance of the point P from the rotation axis, then the surface area swept by the curve segment is da =2πR ds (the surface area of a cylinder of radius R and height ds).

196 90 0. PLANAR CURVES point (x, y) tothex axis) and A =2π y ds =2π C b a (dx ) 2 ( dy ) 2 y(t) + dt. dt dt Example 0.3. Find the area of the surface obtained by revolving one arch of the cycloid x = R(φ sin φ), y = R( cos φ) about the x axis. Solution: The differential of the arc length of the cycloid has been computed in Example 0.. Since y(t) 0 here, the absolute value may be omitted and 2π A =2π yds=2π R( cos φ) 2( cos φ) Rdφ C 0 2π π =8πR 2 sin 3 (φ/2) dφ =6πR 2 sin 3 udu 0 π =6πR 2 ( cos 2 u) sin udu=6πr 2 ( z 2 ) dz 0 =6πR 2 (z z 3 /3) = 64πR2. 3 where the double-angle identity has been used again, sin 2 (φ/2)=( cos φ)/2, and then two successive changes of the integration variable have been done to evaluate the integral, u = φ/2 [0,π] and z = cos u [, ] Exercises. In () (3), find the arc length of the curve. () x =2+3t 2, y = 2t 3 between the points (2, ) and (5, ). (2) x = 3 sin t sin(3t), y = 3 cos t cos(3t), 0 t π. (3) x = t/(+t), y = ln(+t) between the points (0, 0) and (2/3, ln 3). (4) Find the area of the region enclosed by the curve x = a cos 3 t, y = a sin 3 t (the astroid). (5) Find the area of the region enclosed by the curve x = a cos t, y = b sin t (an ellipse). (6) Find the area enclosed by the curve x = t 2 2t, y = t and the y axis. In (7) (0), find the area of a surface generated by rotating the given curve about the specified axis. Sketch the surface. (7) x = a cos 3 t, y = a sin 3 t (about the x axis). (8) x 2 + y 2 = a 2 (about the y axis). (9) x = t 3, y = t 2,0 t (about the x axis). 0

197 69. AREAS AND ARC LENGTHS IN POLAR COORDINATES 9 (0) x = e t t, y =4e t/2,0 t (about the y axis). () Let V be the volume a solid bounded by an axially symmetric surface. Show that V = π C [R(x, y)]2 ds, where C is the curve whose revolution about the symmetry axis gives the boundary surface and R(x, y) is defined in (0.8). Find the volume of the solid bounded by the surface described in Example Areas and Arc Lengths in Polar Coordinates 69.. Area of a Planar Region. Theorem 0.3 (Area of a Planar Region in Polar Coordinates). Let a planar region D be bounded by two rays from the origin θ = a, θ = b and a polar graph r = f(θ), where f(θ) 0, that is, D = {(r, θ) 0 r f(θ), a θ b}. Then the area of D is b A = [f(θ)] 2 dθ. 2 a Proof. Consider a partition of the interval [a, b] by points θ k = a + k θ, k = 0,,..., n, where θ = (b a)/n. Let m k and M k be the minimum and maximum values of f on [θ k,θ k ]. Recall that a continuous function f always attains its maximum and minimum values on a closed interval. The area A k of the planar region bounded by the rays θ = θ k, θ = θ k and the polar graph r = f(θ) is not less than the area of the disk sector with radius r = m k and angle θ and is not greater than the area of the disk sector with radius r = M k and angle θ. The area of a sector of a disk with radius R and angle φ radians is A = 2 R2 φ. Therefore, 2 m2 k θ A k 2 M k 2 θ, and the total area A of the planar region in question satisfies the inequality A L n = n m 2 k θ A n n Mk 2 θ = A U n, A = A k, 2 2 k= k= which is true for any n. Let F (θ) = 2 [f(θ)]2. The function F is continuous on [a, b]. Then 2 m2 k and M 2 2 k are the minimum and maximum values of F on the partition interval [θ k,θ k ]. This shows that the lower and upper bounds, A L n and A U n, are lower and upper sums for the function F on [a, b]. By the definition of the definite integral and k=

198 92 0. PLANAR CURVES integrability of a continuous function (see Calculus I), the upper and lower sums converge to the integral of F over [a, b], A L n b Fdθand a A U n b Fdθas n. The conclusion of the theorem, A = b Fdθ, a a follows from the squeeze principle. Example 0.4. Find the area enclosed by one loop of the four-leaf rose r = cos(2θ). Solution: Note that r = when θ = 0, which is the maximal value of r. The function cos(2θ) has two roots θ = ±π/4 that are the nearest to θ = 0. Hence, one loop corresponds to the interval θ [ π/4,π/4]. The area is π/4 π/4 A = cos 2 (2θ) dθ = 2 π/4 4 = 4 [θ + sin(4θ)] π/4 = π 4 8. π/4 π/4 [ + cos(4θ)] dθ Let D be a planar region that lies between two polar graphs r = f(θ) and r = g(θ) such that f(θ) g(θ) 0ifθ [a, b] and 0 < b a 2π; that is, D is the set of points whose polar coordinates satisfy the inequalities: D = {(r, θ) 0 g(θ) r f(θ),a θ b}. Then the area of D is given by A = 2 b a [f(θ)] 2 dθ 2 b a [g(θ)] 2 dθ = 2 b a ) ([f(θ)] 2 [g(θ)] 2 dθ. Example 0.5. Find the area of a region D bounded by the cardioid r = + sin θ and the circle r =3/2 that lies above the polar axis (in the first and second quadrants). Solution: The polar graphs r = + sin θ = f(θ) and r =3/2 =g(θ) are intersecting when f(θ) = g(θ) or + sinθ = 3/2 or sinθ = /2. Since the region D lies in the first two quadrants, that is, 0 θ π, the values of θ for the points of intersection have to be chosen as θ = π/6 =a and θ = π π/6 =b. Therefore, D = {(r, θ) 3/2 r + sin θ, π/6 θ 5π/6},

199 69. AREAS AND ARC LENGTHS IN POLAR COORDINATES 93 and hence the area of D is A = 2 = 2 = 2 b a b a b a [( + sin θ) ] dθ = 2 b [ sin θ + ( cos(2θ))] dθ 4 2 [ sin θ cos(2θ)] dθ 4 2 a [ sin θ + sin2 θ] dθ = [ 3θ 2 cos θ sin(2θ)] 5π/6 = 9 3 2π π/6 8 Remark. When finding points of intersection of two polar graphs, r = f(θ) and r = g(θ), by solving the equation f(θ) = g(θ), one has to keep in mind that a single point has many representations as described in (0.6). So some of the pairs (f(θ),θ), where θ ranges over solutions of the equation f(θ) = g(θ), may correspond to the same point. To select distinct points, all pairs (f(θ),θ) satisfying the intersection condition can be transformed by means of (0.6) so that r [0, ) and θ [0, 2π). In this range of polar coordinates, there is a one-to-one correspondence between points on a plane and pairs (r, θ) with just one exception when r = 0; all the pairs (0,θ) correspond to the origin of the polar coordinate system Arc Length. Suppose that a curve C is traversed by the point (r, θ) =(f(θ),θ) only once as θ increases from a to b. Choosing θ as a parameter, the curve is described by the parametric equations x = r cos θ, y = r sin θ, where r = f(θ). To find the arc length of C, one has to find the relation between the arc length differential ds and dθ. One has ( dr ) ( dr ) dx = dθ cos θ r sin θ dθ, dy = dθ sin θ + r cos θ dθ Therefore, [( dr ) 2 ( dr ) 2 ] ds 2 =dx 2 + dy 2 = dθ cos θ r sin θ + dθ sin θ + r cos θ dθ 2 [( dr ) 2(cos = 2 θ + sin 2 θ)+r 2 (cos 2 θ + sin θ)] 2 dθ 2 = dθ [( dr dθ ) 2 + r 2] dθ 2.

200 94 0. PLANAR CURVES The arc length of the curve C is b ds b ( dr ) 2 L = ds = C a dθ dθ = r 2 + dθ. a dθ where r = f(θ) and b>a. Example 0.6. Find the length of the cardioid r = + sin θ. Solution: One has ( dr ) 2 r 2 + = ( + sin θ) 2 + (cos θ) 2 = 2( + sin θ), dθ where the trigonometric identity sin 2 θ + cos 2 θ = has been used. The cardioid is traversed once if θ [ π, π]. Therefore, the length is L = 2 π π π/2 + sin θdθ=2 2 + sin θdθ, π/2 since the cardioid is symmetric about the vertical line (the y axis). This integral can be evaluated by the substitution u = + sin θ [0, 2] so that du = cos θdθ, where cos θ = sin 2 θ = (u ) 2 = u(2 u). Hence, L = u du =2 2 2 u(2 u) 0 du = u 2 =8. 2 u Surface Area. If a surface is obtained by rotating a polar graph r = f(θ) about a line, the (0.8) can be used to find the area of the surface where the distance R(x, y) and the arc length differential ds have to be expressed in the polar coordinates with r = f(θ). Example 0.7. Find the area of the surface obtained by rotating the cardioid r = + sin θ about its symmetry axis. Solution: The symmetry axis of the cardioid is the y axis. So the distance from the y axis to a point (x, y) isr(x, y) = x. The surface can be obtained by rotating the part of the cardioid that lies in the fourth and first quadrants, that is, x 0orθ [ π/2,π/2]. Since x = r cos θ, the surface area is π/2 A =2π x ds =2π r cos θ ds dθ dθ =2π C π/2 π/2 r cos θ π/2 r 2 + ( dr ) 2 dθ. dθ

201 69. AREAS AND ARC LENGTHS IN POLAR COORDINATES 95 The derivative ds/dθ has been calculated in the previous example. Therefore, A =2π 2 =2π 2 π/2 π/2 ( + sin θ) 3/2 cos θdθ=2π 2 ( + u)5/2 5/2 = 32π 5, ( + u) 3/2 du where the substitution u = sin θ has been made to evaluate the integral Exercises. In () (4), sketch the curve and find the area that it encloses. () r = 4 cos(2θ) (2) r = a( + cos θ) (3) r =2 cos(2θ) (4) r 2 = 4 cos(2θ) In (5) (7), sketch the curve and find the area of one loop of the curve. (5).r= 9 sin(3θ) (6) r = +2 sin θ (inner loop) (7) r = 2 cos θ sec θ In (8) and (9), find the area of the region that lies inside the first curve and outside the second curve. Sketch the curves. (8) r = 2 sin θ, r= (9) r = 3 cos θ, r= + cos θ In (0) (3), find the area of the region bounded by the curves. Sketch the region. (0) r =2θ, r= θ,θ [0, 2π] () r 2 = sin(2θ),r 2 = cos(2θ) (2) r =2asin θ, r=2bcos θ, a,b>0 (3) r =3+2cosθ, r=3+2sinθ (4) Find the area inside the larger loop and outside the smaller loop of the limaçon r =/2 cos θ. In (5) (7), sketch the curve and find its length. (5) r =2a sin θ (6) r = θ, θ [0, 2π] (7) r = a + cos θ, a In (8) (20), find the area of the surface obtained by rotating the curve about the specified axis. Sketch the surface. (8) r = a>0 about a line through the origin. (9) r =2a cos θ, a>0, (i) about the y axis and (ii) about the x axis. (20) r 2 = cos(2θ) about the polar axis. (2) θ = a and 0 r R, where 0 <a<π/2 and R>0 about the polar axis.

202 96 0. PLANAR CURVES 70. Conic Sections Consider two intersecting lines in space, L and L 2. A surface swept by the line L 2 when it is rotated about the line L is a circular double cone. The line L is the symmetry axis of the cone. The point of intersection of the lines is called the vertex of a cone. Any plane that does not pass through the vertex intersects the cone along a curve. It appears that all such curves fall into three types as shown in Figure 0.4. If the curve of intersection is a loop, then it is an ellipse. If the plane is parallel to the line L 2, then the curve is a parabola. If the plane is parallel to the axis of the cone, then the curve is a hyperbola. The curves of intersection of a plane and a cone are called conic sections, or conics. They have a pure geometrical description, which will be presented here. Remark. A trajectory of any massive object in the solar system (e.g., comet, asteroid, planet) is a conic section that is, a parabola, hyperbola, or ellipse. This fact follows from Newton s Law of Gravity and will be proved in Calculus III Parabolas. A parabola is the set of points in a plane that are equidistant from a fixed point F (called the focus) and a fixed line Figure 0.4. Conic sections are curves that are intersections of a cone with various planes. The shape of a conic section depends on the orientation of the plane relative to the cone symmetry axis.

203 70. CONIC SECTIONS 97 Figure 0.5. Left: Geometrical description of a parabola as a set of points P in a plane that are equidistant from a fixed point F, called the focus, and a fixed line called the directrix (a horizontal line in the figure). Right: A circular paraboloid is the surface obtained by rotating a parabola about the line through its focus and perpendicular to its directrix. (called the directrix). Let P be a point in a plane. Consider the line through P that is perpendicular to the directrix and let Q be the point of their intersection. Then P lies on a parabola if FP = QP. This condition is used to derive the equation of a parabola. A particularly simple equation of a parabola is obtained if the coordinate system is set so that the y axis coincides with the line through the focus and perpendicular to the directrix. The origin O is chosen so that F =(0,p) and hence the parabola contains the origin O, while the directrix is the line y = p parallel to the x axis (the origin is at distance p from F and from the directrix). If P =(x, y), then FP = x 2 +(y p) 2, the point Q has the coordinates (x, p), and PQ = (y + p) 2. An equation of the parabola with focus (0,p) and directrix y = p is FP 2 = PQ 2 = x 2 +(y p) 2 =(y + p) 2 x 2 =4py. In the 6th century, Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. The surface obtained by rotating a parabola about its symmetry axis is called a paraboloid. If a source of light is placed at the focus of a paraboloid mirror, then, after the reflection, the light forms a beam parallel to the symmetry axis. This fact is used to design flashlights, headlights, and

204 98 0. PLANAR CURVES so on. Conversely, a beam of light parallel to the symmetry axis of a paraboloid mirror will be focused to the focus point after the reflection, which is used to design reflecting telescopes Ellipses. An ellipse is the set of points in a plane, the sum of whose distances from two fixed points F and F 2 is a constant. The fixed points are called foci (plural of focus). Let P be a point on a plane. Then P belongs to an ellipse if PF + PF 2 =2a, where a>0 is a constant (the factor 2 is chosen for convenience to be seen later). Evidently, F F 2 < 2a; otherwise, no ellipse exists. A particularly simple equation of an ellipse is obtained when the coordinate system is set so that the foci lie on the x axis and have the coordinates F = ( c, 0) and F 2 = (c, 0), where c < a. Let P =(x, y) be a point in a plane. Then PF = (x + c) 2 + y 2 and PF 2 = (x c) 2 + y 2. The point P is on an ellipse if (0.9) PF + PF 2 =2a PF 2 =2a PF, PF 2 2 =(2a PF ) 2 =4a 2 4a PF + PF 2, 6a 2 PF 2 =(4a 2 + PF 2 PF 2 2 ) 2. These transformations serve only one purpose, that is, to get rid of the square roots. Note that now all the distances are squared. So PF 2 PF 2 2 =(x + c) 2 + y 2 (x c) 2 y 2 =4cx. The substitution of the latter into the condition (0.9) yields 6a 2 [(x+c) 2 +y 2 ]=(4a 2 +4cx) 2 (a 2 c 2 )x 2 +a 2 y 2 = a 2 (a 2 c 2 ). By dividing both sides of this equation by a 2 (a 2 c 2 ), an equation of an ellipse with foci (±a, 0) becomes x 2 a 2 + y2 b 2 =, where b 2 = a 2 c 2 so that a b>0. The ellipse intersects the x axis at (±a, 0) and the y axis at (0, ±b) (called the vertices of an ellipse). The line segment joining the points (±a, 0) is called the major axis. If the foci of an ellipse are located on the y axis, then x and y are swapped in this equation, and the major axis lies on the y axis. This shows that the restriction a b can be dropped in the ellipse equation. In particular, an ellipse becomes a circle of radius a if a = b. One of Kepler s laws is that the orbits of the planets in the solar system are ellipses with the Sun at one focus.

205 70. CONIC SECTIONS 99 Figure 0.6. Left: An ellipse is the set of points in a plane, the sum of whose distances from two fixed points F and F 2 (the foci) is a constant. Right: A circular ellipsoid is the surface obtained by rotating an ellipse about the line through its foci Hyperbolas. A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points F and F 2 (the foci) is a constant. For any point P on a hyperbola, PF PF 2 = ±2a (as the difference of the distances can be negative). Let the foci be at (±c, 0). Following the same procedure used to derive an equation of an ellipse, an equation of a hyperbola with foci (±c, 0) is found to be x 2 a y2 2 b =, 2 where c 2 = a 2 + b 2. The details are left to the reader as an exercise. This equation shows that x 2 /a 2 for any y, that is, x a or x a. A hyperbola therefore has two branches. The branch in x aintersects the x axis at x = a, while the branch in x a does so at x = a. The points (±a, 0) are called vertices. Furthermore, in the asymptotic region x, a hyperbola has slant asymptotes y = ±(b/a)x. Indeed, x 2 y = ±b a =±b x 2 a ) ( a2 x ±b x a2 ± b x 2 a 2x 2 a as x. Here the linearization +u +u/2 has been used to obtain the asymptotic behavior for small u = a 2 /x 2 0.

206 PLANAR CURVES Figure 0.7. Left: A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points F and F 2 (the foci) is a constant. Top right: A circular hyperboloid of one sheet is the surface obtained by rotating a hyperbola about the line through the midpoint of the segment F F 2 and perpendicular to it (the vertical line in the left panel). Bottom right: A circular hyperboloid of two sheets is the surface obtained by rotating a hyperbola about the line through its foci (the horizontal line in the left panel). If the foci of a hyperbola are on the y axis, then, by reversing the roles of x and y, it follows that the hyperbola y 2 a x2 2 b = 2 has foci (0, ±c), where c 2 = a 2 + b 2, vertices (0, ±a), and slant asymptotes y = ±(a/b)x Shifted Conics. Consider a curve defined by a quadratic Cartesian equation Ay 2 + Bx 2 + αy + βx + γ =0.